Calculus 1 - Limit

Contents: Limit of a Function Definition Laws of Limit Limits of Trigonometric Functions Limits of Some Functions Continuity of a Function Theorems Sample Problems Derivative of a Function Applications of Derivative Function of Two or More Variables

Definition

Precise Definition of Limit
Let f be a function defined on an open interval containing a, with the possible exception of a itself (the function f does not need to be defined on a), then $\lim_{x\rightarrow a} f(x)=L$ if for every number Ɛ > 0, exists a number 𝛿 > 0 such that: 0 < |x - a| < 𝛿 implies that |f(x) - L| < Ɛ.

General Definition of Limit
Let f be a function defined on an open interval containing a, with the possible exception of a itself (the function f does not need to be defined on a), then $\lim_{x\rightarrow a} f(x)=L$ if f(x) can be made as close to L by making x to be sufficiently close to a, from both right and left side: $\lim_{x\rightarrow a^+} f(x)=\lim_{x\rightarrow a^-} f(x)=L$.

Laws of Limit

$\begin{align*} \text{1. }&\lim_{x\rightarrow a}c=c\\ \text{2. }&\lim_{x\rightarrow a}x=a\\ \text{3. }&\lim_{x\rightarrow a}(f(x)\pm g(x))=\lim_{x\rightarrow a}f(x)\pm \lim_{x\rightarrow a}g(x)\\ \text{4. }&\lim_{x\rightarrow a}(f(x).g(x))=\lim_{x\rightarrow a}f(x). \lim_{x\rightarrow a}g(x)\\ \text{5. }&\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)},\,g(x)\neq0\\ \text{6. }&\lim_{x\rightarrow a} c.f(x)=c.\lim_{x\rightarrow a} f(x)\\ \text{7. }&\lim_{x\rightarrow a} \left [ f(x) \right ]^{n}= \left [ \lim_{x\rightarrow a} f(x) \right ] ^{n},n\,\epsilon\,\mathbb{R}\\ \end{align*}$

Limits of Trigonometric Functions

$\begin{align*} \text{8. }&\lim_{x\rightarrow 0}\frac{\sin{ax}}{bx}=\lim_{x\rightarrow 0}\frac{ax}{\sin{bx}}=\frac{a}{b}\\ \text{9. }&\lim_{x\rightarrow 0}\frac{\tan{ax}}{bx}=\lim_{x\rightarrow 0}\frac{ax}{\tan{bx}}=\frac{a}{b}\\ \text{10. }&\lim_{x\rightarrow 0}\frac{\sin{ax}}{\tan{bx}}=\lim_{x\rightarrow 0}\frac{\tan{ax}}{\sin{bx}}=\frac{a}{b}\\ \text{11. }&\lim_{x\rightarrow 0}\frac{\sin{ax}}{\sin{bx}}=\lim_{x\rightarrow 0}\frac{\tan{ax}}{\tan{bx}}=\frac{a}{b}\\ \end{align*}$

Limits of Some Functions

$\begin{align*} \small\text{12. }&\lim_{x\rightarrow \infty}\left ( 1+\frac{k}{x}\right )^x = e^k\\ \small\text{13. }&\lim_{x\rightarrow \infty } (\sqrt[n]{ax^{n}+bx^{n-1}+...}-\sqrt[n]{px^{n}+qx^{n-1}+...})=\left\{\begin{matrix} \infty &,a>p\\ \frac{b-p}{2\sqrt{a}}&,a=p\\ 0&,a<p \end{matrix} \right. \end{align*}$

Continuity of a Function

Let f be a function defined on an open interval containing all values of x close to a. The function f is continuous at a if $\lim_{x\rightarrow a} f(x)=f(a)$.

Suppose f and g are both function that is continuous at a. Then, the following functions are also continuous at a:

• f(x) ± g(x)
• f(x) . g(x)
• f(x) / g(x)
• c . f(x)

The function f is:

• continuous from the right at a if $\lim_{x\rightarrow a^+} f(x)=f(a)$
• continuous from the left at a if $\lim_{x\rightarrow a^-} f(x)=f(a)$
• continuous on an open interval (a,b) if f is continuous on every number k where a < k < b.
• continuous on a closed interval [a,b] if f is continuous on every number k where a ≤ k ≤ b.

There are three types of discontinuity:

• Jump Discontinuity:
  
  Occurs when: $\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)$
• Infinite Discontinuity:
  
  Occurs when the function approaches infinity or negative infinity.
• Removable Discontinuity
  
  Occurs when: $\lim_{x\rightarrow a} f(x)=f(a)$, or f(a) doesn't exist.

Theorems

Squeeze Theorem
Suppose that f(x) ≤ g(x) ≤ h(x) for all x in an open interval containing a, except possibly at a, and $\lim_{x\rightarrow a} f(x)=\lim_{x\rightarrow a} h(x)=L$. So,$\lim_{x\rightarrow a} g(x)=L$.

Intermediate Value Theorem
Suppose that f(x) is a function continuous on the interval [a,b], and C is such that f(a) < C < f(b). Then, there exists a number c such that f(c) = C.

Sample Problems

Problem 1: Finding Limit
Find the limit of the given function without the L'Hôpital's rule:

$\begin{align*} \text{A. } &\lim_{x\rightarrow 1} \frac{x^3+x^2-x-1}{x^4-x^3+x-1}\\ \text{B. } &\lim_{x\rightarrow 4} \frac{2-\sqrt{x}}{8-\sqrt{x^3}}\\ \text{C. } &\lim_{x\rightarrow 2} \left ( \frac{1}{2-x}-\frac{12}{8-x^3} \right )\\ \text{D. } &\lim_{x\rightarrow 0} \frac{x}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}\\ \text{E. } &\lim_{x\rightarrow 0} \frac{3^x-1}{27^x-1}\\ \text{F. } &\lim_{x\rightarrow 2} \frac{x^2+2x+4}{x^2-4x+4}\\ \text{G. } &\lim_{x\rightarrow 0} (1+4x)^{\frac{1}{x}}\\ \text{H. } &\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-(x-4)\\ \text{I. } &\lim_{x\rightarrow \infty} \frac{5x^4-32x}{\sqrt{4x^8+12x^7+9x^6+32}}\\ \text{J. } &\lim_{x\rightarrow 2} \frac{x^2-4}{x^2+2x+3}\\ \end{align*}$

Sub-Problem A
Since substituting x = 1 results in an indeterminate form 0/0, there is a common factor between the numerator dan denominator that needs to be removed. In this case, it is (x-1):
$\begin{align*} &\lim_{x\rightarrow 1} \frac{x^3+x^2-x-1}{x^4-x^3+x-1}\\&=\lim_{x\rightarrow 1} \frac{(x-1)(x^2+2x+1)}{(x-1)(x^3+1)}\\ &=2\\ \end{align*}$

Sub-Problem B
Let us define a variable v such that v² = x. Since this is also becomes 0/0 when x = 4 is substituted, there must be a common factor. Remove the common factor in both numerator and denominator:
$\begin{align*} &\lim_{x\rightarrow 4} \frac{2-\sqrt{x}}{8-\sqrt{x^3}} \\ &=\lim_{v\rightarrow 2} \frac{(2-v)}{(8-v^3)}\\ &=\lim_{v\rightarrow 2} \frac{(2-v)}{(2-v)(4+2v+v^2)}\\ &=\lim_{v\rightarrow 2} \frac{1}{(4+2v+v^2)}\\ &=\frac{1}{12} \end{align*}$

Sub-Problem C
Let us start by combining the two fractions into one, then remove common factors like the previous problems:
$\begin{align*} &\lim_{x\rightarrow 2} \left ( \frac{1}{2-x}-\frac{12}{8-x^3} \right )\\&= \lim_{x\rightarrow 2} \frac{(4+2x+x^2)-12}{(8-x^3)} \\ &= \lim_{x\rightarrow 2} \frac{x^2+2x-8}{(8-x^3)} \\ &= \lim_{x\rightarrow 2} \frac{(x+4)(x-2)}{(x^2+2x+4)(2-x)} \\ &= \lim_{x\rightarrow 2} -\frac{(x+4)}{(x^2+2x+4)} \\ &=-\frac{1}{2} \end{align*}$

Sub-Problem D
Using the form (a³ - b³) = (a - b)(a² + ab + b²), we will substitute the variables, so that:
$\begin{align*} a&=\sqrt[3]{1+x}\\ b&=\sqrt[3]{1-x} \end{align*}$

$\small \begin{align*} &\lim_{x\rightarrow 0} \frac{x}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}\\ &=\lim_{x\rightarrow 0} \frac{x}{a-b}\\ &=\lim_{x\rightarrow 0} \left ( \frac{x}{a-b} \times \frac{\left( a^2+b^2+ab\right)}{\left( a^2+b^2+ab\right)}\right )\\ &=\lim_{x\rightarrow 0} \frac{x( a^2+b^2+ab)}{a^3-b^3} \\ &=\lim_{x\rightarrow 0} \frac{x\left( \sqrt[3]{(1+x)^2}+\sqrt[3]{(1-x)^2}+\sqrt[3]{(1-x)(1+x)}\right)}{(1+x)-(1-x)}\\ &=\lim_{x\rightarrow 0} \frac{x\left( \sqrt[3]{(1+x)^2}+\sqrt[3]{(1-x)^2}+\sqrt[3]{(1-x)(1+x)}\right)}{2x}\\ &= \frac{(1+1+1)}{2}\\ &= \frac{3}{2} \end{align*}$

Sub-Problem E
Let us define a variable v such that v = 3^x. Since this is also becomes 0/0 form when x = 2 is substituted, there must be a common factor. Remove the common factor in both numerator and denominator:
$\small \begin{align*} &\lim_{x\rightarrow 0} \frac{3^x-1}{27^x-1}\\&=\lim_{x\rightarrow 0}\frac{3^x-1}{3^{3x}-1}\\ &=\lim_{v\rightarrow 1}\frac{v-1}{v^{3}-1}\\ &=\lim_{v\rightarrow 1}\frac{v-1}{(v-1)(v^2+v+1)}\\ &=\lim_{v\rightarrow 1}\frac{1}{(v^2+v+1)}\\ &=\frac{1}{3} \end{align*}$

Sub-Problem F
$\small \begin{align*} \lim_{x\rightarrow 2} \frac{x^2+2x+4}{x^2-4x+4}&=\frac{12}{0}\\ &=\infty \end{align*}$

Sub-Problem G
Recall that:
$\lim_{x\rightarrow \infty} (1+\frac{k}{x})^{x}=e^k\\ $
Let us define a variable v such that v = 1/x. So:
$\begin{align*} &\lim_{x\rightarrow 0} (1+4x)^{\frac{1}{x}}\\&=\lim_{v\rightarrow \infty} (1+\frac{4}{v})^{v}\\ &=e^4 \end{align*}$

Sub-Problem H
Since substituting x = infinity yields an indeterminate form, a different approach is needed. Let us start by converting (x-4):
$\small\begin{align*} &\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-(x-4)\\ &=\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-\sqrt{(x-4)^2}\\ &=\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-\sqrt{x^2-8x+16}\\ &=\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-\sqrt{x^2-8x+16}\times \frac{\sqrt{x^2+6x-9}+\sqrt{x^2-8x+16}}{\sqrt{x^2+6x-9}+\sqrt{x^2-8x+16}}\\ &=\lim_{x\rightarrow \infty} \frac{(x^2+6x-9)-(x^2-8x+16)}{\sqrt{x^2+6x-9}+\sqrt{x^2-8x+16}}\\ &=\lim_{x\rightarrow \infty} \frac{14x-25}{\sqrt{x^2+6x-9}+\sqrt{x^2-8x+16}}\\ &=\lim_{x\rightarrow \infty} \frac{14-\frac{25}{x}}{\sqrt{1+\frac{6}{x}-\frac{9}{x^2}}+\sqrt{1-\frac{8}{x}+\frac{16}{x^2}}}\\ &=\frac{14-0}{\sqrt{1+0-0}+\sqrt{1-0+0}}\\ &=7 \end{align*}$

Or, just use Formula 13 (Limit):
$\begin{align*} &\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-(x-4)\\ &=\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-\sqrt{(x-4)^2}\\ &=\lim_{x\rightarrow \infty} \sqrt{x^2+6x-9}-\sqrt{x^2-8x+16}\\ &=\frac{6-(-8)}{2\sqrt{1}}\\ &=7 \end{align*}$

Sub-Problem I
$\small \begin{align*} &\lim_{x\rightarrow \infty} \frac{5x^4-32x}{\sqrt{4x^8+12x^7+9x^6+32}}\\ &=\lim_{x\rightarrow \infty} \sqrt{\frac{25x^8-320x^5+1024x^2}{4x^8+12x^7+9x^6+32}}\\ &=\sqrt{\lim_{x\rightarrow \infty}\frac{25x^8-320x^5+1024x^2}{4x^8+12x^7+9x^6+32} }\\ &=\sqrt{\lim_{x\rightarrow \infty}\frac{25-\frac{320}{x^3}+\frac{1024}{x^6}}{4 +\frac{12}{x}+\frac{9}{x^2}+\frac{32}{x^8}} }\\ &=\sqrt{\frac{25}{4}}\\ &=\frac{5}{2} \end{align*}$

Sub-Problem J
$\begin{align*} \lim_{x\rightarrow 2} \frac{x^2-4}{x^2+3x+2}&=\frac{0}{12}\\ &=0 \end{align*}$

Problem 2: Finding Limit & Squeeze Theorem
Find the limit of the given function without the L'Hôpital's rule:
$\begin{align*} \text{A. } &\lim_{x\rightarrow 0} \frac{1-\cos{2x}}{3x \sin{4x}}\\ \text{B. } &\lim_{x\rightarrow \infty} \frac{\csc^{-1}{x}}{x^2-x}\\ \text{C. } &\lim_{x\rightarrow \infty} \frac{\sin{x}}{x}\\ \text{D. } &\lim_{x\rightarrow 0} \frac{\tan{2x}}{\sqrt{1-\sin{x}}-\sqrt{1+\sin{x}}}\\ \text{E. } &\lim_{x\rightarrow 0} \frac{\sin{3x}+\tan{4x}}{\sin{4x}-\tan{3x}}\\ \text{F. } &\lim_{x\rightarrow 0} \frac{\cos{16x}-\cos{8x}}{\cos{4x}-\cos{x}}\\ \text{G. } &\lim_{x\rightarrow 0} \frac{\sin^{-1}\left(x\right)}{x}\\ \text{H. } &\lim_{x\rightarrow \infty} \frac{\sin^2{(2x-1)-5x}}{1-5x}\\ \text{I. } &\lim_{x\rightarrow 0} \frac{\sin{x}-\tan{x}}{x^3}\\ \text{J. } &\lim_{x\rightarrow 0} \sin{x}\cos{\frac{1}{x}}\\ \end{align*}$

Sub-Problem A
Using the following identity:
$\begin{align*} &\cos{2x}=\cos^2{x}-\sin^2{x}\\ &\sin^2{x}+\cos^2{x}=1\\ \end{align*}$

$\begin{align*} &\lim_{x\rightarrow 0} \frac{1-\cos{2x}}{3x \sin{4x}}\\ &=\lim_{x\rightarrow 0} \frac{1-(\cos^2{x}-\sin^2{x})}{3x \sin{4x}}\\ &=\lim_{x\rightarrow 0} \frac{1-\cos^2{x}+\sin^2{x}}{3x \sin{4x}}\\ &=\lim_{x\rightarrow 0} \frac{2\sin^2{x}}{3x \sin{4x}}\\ &=\lim_{x\rightarrow 0} \frac{2\sin{x}\sin{x}}{3x \sin{4x}}\\ &=2 \times \frac{1}{3} \times \frac{1}{4}\\ &=\frac{1}{6} \end{align*}$

Sub-Problem B
Warning: $\csc^{-1}(x) \neq \frac{1}{\csc(x)}$
The value of csc^-1(x) ranges from -$\frac{\pi}{2}$ to $\frac{\pi}{2}$:
$\begin{align*} -\frac{\pi}{2}\leq&\csc^{-1}{x}\leq \frac{\pi}{2}\\ -\frac{\pi}{2(x^2-x)}\leq&\frac{\csc^{-1}{x}}{x^2-x}\leq \frac{\frac{\pi}{2}}{x^2-x}\\ \end{align*}$

Now, we know that:
$\begin{align*} &\lim_{x\rightarrow \infty} -\frac{\pi}{2(x^2-x)}=0\\ &\lim_{x\rightarrow \infty} \frac{\pi}{2(x^2-x)}=0\\ \end{align*}$

By Squeeze Theorem, then:
$\begin{align*} \lim_{x\rightarrow \infty} \frac{\csc^{-1}{x}}{x^2-x}=0 \end{align*}$

Sub-Problem C
The value of sin(x) ranges from -1 to 1:
$\begin{align*} -1\leq&\sin{x}\leq 1\\ -\frac{1}{x}\leq&\frac{\sin{x}}{x} \leq \frac{1}{x}\\ \end{align*}$

Now, we know that:
$\begin{align*} &\lim_{x\rightarrow \infty} -\frac{1}{x}=0\\ &\lim_{x\rightarrow \infty} \frac{1}{x}=0\\ \end{align*}$

By Squeeze Theorem, then:
$\begin{align*} \lim_{x\rightarrow \infty} \frac{\sin{x}}{x}=0 \end{align*}$

Sub-Problem D
$\small \begin{align*} &\lim_{x\rightarrow 0} \frac{\tan{2x}}{\sqrt{1-\sin{x}}-\sqrt{1+\sin{x}}}\\ &=\lim_{x\rightarrow 0} \frac{\tan{2x}}{\sqrt{1-\sin{x}}-\sqrt{1+\sin{x}}}\times \frac{\sqrt{1-\sin{x}}+\sqrt{1+\sin{x}}}{\sqrt{1-\sin{x}}+\sqrt{1+\sin{x}}}\\ &=\lim_{x\rightarrow 0} \frac{\tan{2x}}{-2\sin{x}} (\sqrt{1-\sin{x}}+\sqrt{1+\sin{x}})\\ &=-2 \end{align*}$

Sub-Problem E
$\small \begin{align*} &\lim_{x\rightarrow 0} \frac{\sin{3x}+\tan{4x}}{\tan{3x}-\sin{4x}}\\ &=\lim_{x\rightarrow 0} \frac{\sin{3x}+\tan{4x}}{\tan{3x}-\sin{4x}}\times \frac{\frac{1}{\sin{x}}}{\frac{1}{\sin{x}}}\\ &=\lim_{x\rightarrow 0} \frac{\frac{\sin{3x}}{\sin{x}}+\frac{\tan{4x}}{\sin{x}}}{\frac{\tan{3x}}{\sin{x}}-\frac{\sin{4x}}{\sin{x}}}\\ &=\frac{3+4}{3-4}\\ &=-7 \end{align*}$

Sub-Problem F
Using the following identity:
$\cos{A}-\cos{B}=-2 \sin{\left ( \frac{A+B}{2} \right )} \sin{\left ( \frac{A-B}{2} \right )}$

$\begin{align*} &\lim_{x\rightarrow 0} \frac{\cos{16x}-\cos{8x}}{\cos{4x}-\cos{x}}\\ &=\lim_{x\rightarrow 0} \frac{-2 \times \sin{12x} \times \sin{4x}}{-2 \times \sin{\frac{5}{2}x \times \sin{\frac{3}{2}x}}}\\ &=\lim_{x\rightarrow 0} \frac{-2 \times \sin{12x} \times \sin{4x}}{-2 \times \sin{\frac{5}{2}x \times \sin{\frac{3}{2}x}}}\\ &=\frac{24}{5}\times\frac{8}{3}\\ &=\frac{64}{5} \end{align*}$

Sub-Problem G
Warning: $\sin^{-1}(x) \neq \frac{1}{\sin(x)}$
Let us define a new variable u, where:
$\begin{align*} u&=\sin^{-1}(x)\\ \sin{u}&=\sin(\sin^{-1}(x))\\ \sin{u} &= x \end{align*}$

From the above, we can see that if x approaches 0, then sin(u) would approach 0 too. So:
$\begin{align*} \lim_{x\rightarrow 0} \frac{\sin^{-1}(x)}{x}&= \lim_{u\rightarrow 0} \frac{u}{\sin{u}}\\ &=1\\ \end{align*}$

Sub-Problem H
The range of sin² is from 0 to 1:
$\begin{align*} -1\leq&\sin{(2x-1)}\leq 1\\ 0\leq&\sin^2{(2x-1)}\leq 1\\ \frac{0}{x}\leq&\frac{\sin^2{(2x-1)}}{x}\leq \frac{1}{x}\\ \end{align*}$

Now, we know that:
$\begin{align*} &\lim_{x\rightarrow \infty} \frac{0}{x}=0\\ &\lim_{x\rightarrow \infty} \frac{1}{x}=0\\ \end{align*}$

By Squeeze Theorem, then:
$\begin{align*} \lim_{x\rightarrow \infty} \frac{\sin{2x-1}}{x}=0 \end{align*}$

So, the limit can become:
$\begin{align*} &\lim_{x\rightarrow \infty} \frac{\sin^2{(2x-1)-5x}}{1-5x}\\ &= \lim_{x\rightarrow \infty} \frac{\sin^2{(2x-1)-5x}}{1-5x} \times \frac{\frac{1}{x}}{\frac{1}{x}}\\ &= \lim_{x\rightarrow \infty} \frac{\frac{\sin^2{(2x-1)}}{x}-5}{\frac{1}{x}-5}\\ &= \frac{0-5}{0-5}\\ &= 1 \end{align*}$

Sub-Problem I
Using the following identity:
$\begin{align*} &\sin^2{x}+\cos^2{x}=1\\ \end{align*}$

$\begin{align*} &\lim_{x\rightarrow 0} \frac{\sin{x}-\tan{x}}{x^3}\\ &= \lim_{x\rightarrow 0} \frac{\sin{x}-\frac{\sin{x}}{\cos{x}}}{x^3}\\ &= \lim_{x\rightarrow 0} \frac{\sin{x}(1-\frac{1}{\cos{x}})}{x^3}\\ &= \lim_{x\rightarrow 0} \frac{1-\frac{1}{\cos{x}}}{x^2}\\ &= \lim_{x\rightarrow 0} \frac{\cos{x}-1}{x^2 \cos{x}}\times\frac{\cos{x}+1}{\cos{x}+1}\\ &= \lim_{x\rightarrow 0} \frac{\cos^2{x}-1}{x^2 \cos{x}(\cos{x}+1)}\\ &= \lim_{x\rightarrow 0} \frac{-\sin^2{x}}{x^2 \cos{x}(\cos{x}+1)}\\ &= \lim_{x\rightarrow 0} \frac{-1}{\cos{x}(\cos{x}+1)}\\ &= -\frac{1}{2} \end{align*}$

Sub-Problem J
The range of cos is from -1 to 1:
$\begin{align*} -1\leq&\cos{\frac{1}{x}}\leq 1\\ -\sin{x}\leq&\sin{x}\cos{\frac{1}{x}}\leq\sin{x}\\ \end{align*}$

Now, we know that:
$\begin{align*} &\lim_{x\rightarrow 0} -\sin{x}=0\\ &\lim_{x\rightarrow 0} \sin{x}=0\\ \end{align*}$

By Squeeze Theorem, then:
$\lim_{x\rightarrow 0} \sin{x}\cos{\frac{1}{x}}=0\\ $

Problem 3: Identifying Continuity
Given f(x) and a, show whether the given functions are continuous at a.

$\\ \text{A. }a=3\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} 3x-4 & \text{, } x<3 \\ -3x+14& \text{, } x\geq 3 \end{cases}\\ \text{B. }a=1\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} \frac{x^3-1}{x} & \text{, } x\leq1 \\ x^2+2x-3& \text{, } x> 1 \end{cases}\\ \text{C. }a=0\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} \tan{x} \cos{\frac{1}{x}} & \text{, } x\neq0 \\ 0& \text{, } x=0 \end{cases}\\ \text{D. }a=k\\ \text{ }\,\text{ }\,\text{ }f(x)=\frac{1}{x-k}\\ \text{E. } a=1\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} x^2-2x+3 & \text{, } x \neq 1\\ 3& \text{, } x = 1 \end{cases}\\ $

The function f is continuous at a, if $\lim_{x\rightarrow a} f(x) = f(a)$.

Sub-Problem A
Check f(3):
$\begin{align*} f(3)&=-3(3)+14\\ &=-9+14\\ &=5 \end{align*}$

Checking the left and right side limit:
$\begin{align*} &\lim_{x\rightarrow 3^-} 3x-4 = 5\\ &\lim_{x\rightarrow 3^+} -3x+14 = 5\\ \end{align*}$

Since both left and right side limit are the same, we know that:
$\lim_{x\rightarrow 3^-} f(x) = \lim_{x\rightarrow 3^+} f(x) = \lim_{x\rightarrow 3} f(x) = 5$

Since $\lim_{x\rightarrow 3} f(x) = f(a)$, the function f is continuous at a = 3.

Sub-Problem B
Check f(1):
$\begin{align*} f(1)&=\frac{1^3-1}{1}\\ &=0 \end{align*}$

Checking the left and right side limit:
$\begin{align*} &\lim_{x\rightarrow 1^-} \frac{x^3-1}{1} = 0\\ &\lim_{x\rightarrow 1^+} x^2+2x-3 = 0\\ \end{align*}$

Since both left and right side limit are the same, we know that:
$\lim_{x\rightarrow 1^-} f(x) = \lim_{x\rightarrow 1^+} f(x) = \lim_{x\rightarrow 1} f(x) = 0$

Since $\lim_{x\rightarrow 1} f(x) = f(a)$, the function f is continuous at a = 1.

Sub-Problem C
Check f(0):
$\begin{align*} f(0)&=0 \end{align*}$

In order to check the limit of the function when it is not equal to zero, Squeeze Theorem can be used. The value of cos ranges from -1 to 1:
$\begin{align*} -1\leq&\cos{\left (\frac{1}{x} \right)}\leq 1\\ -\tan{x}\leq&\tan{x}\cos{\left (\frac{1}{x} \right)}\leq \tan{x}\\ \end{align*}$

The limit of -tan(x) and tan(x):
$\begin{align*} &\lim_{x\rightarrow 0} -\tan{x}=0\\ &\lim_{x\rightarrow 0} \tan{x}=0\\ \end{align*}$

By Squeeze Theorem, then:
$\begin{align*} &\lim_{x\rightarrow 0} \tan{x}\cos{\left (\frac{1}{x} \right)}=0\\ \end{align*}$

By the definition of limit, and since we know the limit of a function, then we also know both left and right side limit:
$\begin{align*} &\lim_{x\rightarrow 0} \tan{x}\cos{\left (\frac{1}{x} \right)}\\ &=\lim_{x\rightarrow 0^-} \tan{x}\cos{\left (\frac{1}{x} \right)}\\ &=\lim_{x\rightarrow 0^+} \tan{x}\cos{\left (\frac{1}{x} \right)}\\ &=\lim_{x\rightarrow 0} f(x)\\ \end{align*}$

Since $\lim_{x\rightarrow 0} f(x) = f(0)$, f is continuous at a = 0.

Sub-Problem D
Check f(k):
$\begin{align*} f(k)&=\frac{1}{k-k}\\ &=\text{Undefined} \end{align*}$

Checking the left and right side limit:
$\begin{align*} &\lim_{x\rightarrow k^-} \frac{1}{x-k}= -\infty\\ &\lim_{x\rightarrow k^+} \frac{1}{x-k}= \infty\\ \end{align*}$

Since the left and right side limit are inequal, it does not have a limit at x = k. So, the function f is not continuous at x = k.

Sub-Problem E
Check f(1):
$\begin{align*} f(1)&=3 \end{align*}$

Checking the left and right side limit:
$\begin{align*} &\lim_{x\rightarrow 1^-} x^2-2x+3= 2\\ &\lim_{x\rightarrow 1^+} x^2-2x+3= 2\\ \end{align*}$

Since both left and right side limit are the same, we know that:
$\lim_{x\rightarrow 1^-} f(x) = \lim_{x\rightarrow 1^+} f(x) = \lim_{x\rightarrow 1} f(x) = 2$

Since $\lim_{x\rightarrow 1} f(x) \neq f(1)$, f is not continuous at a = 1.

Problem 4: Identifying Continuity
Given f(x) and a, find the constants (c, d) so that the given functions are continuous at a.

$\text{A. }a=0\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} \frac{\tan^2{x}}{1-\cos{x}} & \text{, } x \neq 0\\ c& \text{, } x = 0\\ \end{cases}\\ \text{B. }a=-3\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} x+5& \text{, } x \leq-3\\ c-2x& \text{, } x > -3\\ \end{cases}\\ \text{C. }a=2\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} x^3& \text{, } x \leq 2\\ \frac{c}{x}& \text{, } x > 2\\ \end{cases}\\ \text{D. }a=0\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} \frac{\tan{(dx-x^2)}}{cx}& \text{, } x < 0\\ 2 & \text{, } x = 0\\ -x^2-cx+2 & \text{, } x > 0\\ \end{cases}\\ \text{E. }a=\frac{\pi}{2}\\ \text{ }\,\text{ }\,\text{ }f(x)=\begin{cases} c-x& \text{, } x <\frac{\pi}{2}\\ c \cos{x}& \text{, } x \geq \frac{\pi}{2}\\ \end{cases}\\ $

Sub-Problem A
Check f(0):
$\begin{align*} f(0)&=c \end{align*}$

Since f is continuous on a = 0, it is known that:
$\lim_{x\rightarrow 0} f(x) = f(0)$

Which means:
$\begin{align*} \lim_{x\rightarrow 0} f(x) &= f(0)\\ \lim_{x\rightarrow 0^-} f(x) &= f(0)\\ \lim_{x\rightarrow 0^+} f(x) &= f(0)\\ \end{align*}$

Since 0^- and 0⁺ are in the same domain (not equal to zero), we can just use one of the limits:
$\begin{align*} \lim_{x\rightarrow 0^-} f(x) &= f(0)\\ \lim_{x\rightarrow 0^-} \frac{\tan^2{x}}{1-\cos{x}}&=c\\ \lim_{x\rightarrow 0^-} \frac{\tan^2{x}(1+\cos{x})}{1-\cos^2{x}}&=c\\ \lim_{x\rightarrow 0^-} \frac{\tan^2{x}(1+\cos{x})}{\sin^2{x}}&=c\\ \lim_{x\rightarrow 0^-} \frac{(1+\cos{x})}{\cos^2{x}}&=c\\ c&=2 \end{align*}$

The value of c is 2.

Sub-Problem B
Check f(-3):
$\begin{align*} f(-3)&=(-3)+5\\ &=2 \end{align*}$

Since f is continuous on a = -3, it is known that:
$\lim_{x\rightarrow -3} f(x) = f(-3)$

Which means:
$\begin{align*} \lim_{x\rightarrow -3} f(x) &= f(-3)\\ \lim_{x\rightarrow -3^-} f(x) &= f(-3)\\ \lim_{x\rightarrow -3^+} f(x) &= f(-3)\\ \end{align*}$

Then:
$\begin{align*} \lim_{x\rightarrow -3^-} f(x) &= \lim_{x\rightarrow -3^+} f(x)\\ \lim_{x\rightarrow -3-} x+5&= \lim_{x\rightarrow -3^+} c-2x\\ 2&=c+6\\ c&=-4 \end{align*}$

The value of c is -4.

Sub-Problem C
Check f(2):
$\begin{align*} f(2)&=2^3\\ &=8 \end{align*}$

Since f is continuous on a = 2, it is known that:
$\lim_{x\rightarrow 2} f(x) = f(2)$

Which means:
$\begin{align*} \lim_{x\rightarrow 2} f(x) &= f(2)\\ \lim_{x\rightarrow 2^-} f(x) &= f(2)\\ \lim_{x\rightarrow 2^+} f(x) &= f(2)\\ \end{align*}$

Then:
$\begin{align*} \lim_{x\rightarrow 2^-} f(x) &= \lim_{x\rightarrow 2^+} f(x)\\ \lim_{x\rightarrow 2^-} x^3&= \lim_{x\rightarrow 2^+} \frac{c}{x}\\ 8&=\frac{c}{2}\\ c&=16 \end{align*}$

The value of c is 16.

Sub-Problem D
Check f(0):
$\begin{align*} f(0)&=2\\ \end{align*}$

Since f is continuous on a = 0, it is known that:
$\lim_{x\rightarrow 0} f(x) = f(0)$

Which means:
$\begin{align*} \lim_{x\rightarrow 0} f(x) &= f(0)\\ \lim_{x\rightarrow 0^-} f(x) &= f(0)\\ \lim_{x\rightarrow 0^+} f(x) &= f(0)\\ \end{align*}$

Then:
$\begin{align*} \lim_{x\rightarrow 0^+} f(x) &= f(0)\\ \lim_{x\rightarrow 0^+} -x^2-cx+2c&=2\\ 2c&=2\\ c&=1 \end{align*}$

Finding the value of d. Let us define a variable m so that m = d-x, and using the following identity:
$\tan{(A\pm B)}=\frac{\tan{A}\pm\tan{B}}{1\mp\tan{A}\tan{B}}\\ $

$\begin{align*} \lim_{x\rightarrow 0^-} f(x) &= f(0)\\ \lim_{x\rightarrow 0^-}\frac{\tan{(dx-x^2)}}{cx} &= 2\\ \lim_{x\rightarrow 0^-}\frac{\tan{dx}-\tan{x^2}}{x(1 +\tan{dx}\tan{x^2})} &= 2\\ \left (\lim_{x\rightarrow 0^-}\frac{\tan{dx}}{x}-\lim_{x\rightarrow 0^-} \frac{\tan{x^2}}{x}\right )\times\lim_{x\rightarrow 0^-}\frac{1}{(1 +\tan{dx}\tan{x^2})} &= 2\\ \left (\lim_{x\rightarrow 0^-} \frac{\tan{dx}}{x}-\lim_{x\rightarrow 0^-}\frac{2x \sec{x^2}}{1}\right )\times\lim_{x\rightarrow 0^-} \frac{1}{(1 +\tan{dx}\tan{x^2})}&=2\\ \frac{d-0}{1+0}&=2\\ d&=2 \end{align*}$

The value of c is 1, and d is 2.

Sub-Problem E
Check f(π/2):
$\begin{align*} f\left(\frac{\pi}{2}\right)&=c \cos\left(\frac{\pi}{2}\right)\\ &=0 \end{align*}$

Since f is continuous on a = π/2, it is known that:
$\lim_{x\rightarrow \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)$

Which means:
$\begin{align*} \lim_{x\rightarrow \frac{\pi}{2}} f(x) &= f\left(\frac{\pi}{2}\right)\\ \lim_{x\rightarrow \frac{\pi}{2}^-} f(x) &= f\left(\frac{\pi}{2}\right)\\ \lim_{x\rightarrow \frac{\pi}{2}^+} f(x) &= f\left(\frac{\pi}{2}\right)\\ \end{align*}$

Then:
$\begin{align*} \lim_{x\rightarrow \frac{\pi}{2}^-} f(x) &= \lim_{x\rightarrow \frac{\pi}{2}^+} f(x)\\ \lim_{x\rightarrow \frac{\pi}{2}^-} c-x &= \lim_{x\rightarrow \frac{\pi}{2}^+} c \cos{x}\\ c - \frac{\pi}{2} &= 0\\ c &= \frac{\pi}{2} \end{align*}$

The value of c is π/2.

Problem 5: Identifying Continuity
Given f(x), find the intervals where f(x) is continuous.

$\text{A. }f(x)=x^4+4x^2+1\\ \text{B. }f(x)=\frac{\cos{x}}{x}\\ \text{C. }f(x)=\sqrt{\frac{x^2-7x+6}{-x^2+6x-8}}\\ \text{D. }f(x)=\ln{\sin{x}}\\ \text{E. }f(x)=\csc^2{x}\\ $

Sub-Problem A:
Since it is a regular polynomial, f(x) is continuous on (-∞,∞)

Sub-Problem B:
Since the denominator of a fraction cannot be zero, f(x) isn't continuous at x=0. So, f(x) is continuous on x, such that x ≠ 0.

Sub-Problem C:
Since square root can't be negative, we must find the value(s) of x that makes the fraction inside of the root negative. If one of the numerator or denominator is negative, the whole fraction part will be a negative. Let's check the numerator part first:

The numerator part can actually be factorized, so that it becomes (x-6)(x-1). We can see that (x-6)(x-1) becomes negative, if 1 < x < 6. Or:


The denominator part can also be factorized, so that it becomes -1(x-2)(x-4). We can see that -1(x-2)(x-4) becomes negative, if x < 2 or x > 4. Or:


The denominator part cannot be zero, so obviously x can't be 2 or 4.

If we combine both, we can see that the whole fraction will be positive on (1,2) and (2,4). Which means, f(x) is continuous on (1,2) and (2,4)

Sub-Problem D:
Logarithm cannot be zero, or negative. Let k be an integer. sin(x) will always be positive on (2kπ, (2k+1)π)
f(x) is continuous on (2kπ, (2k+1)π).

Sub-Problem E:
Let k be an integer. f(x) is continuous on x, such that x ≠ nπ, where n is an integer.

Problem 6: Identifying Continuity
Check the validity of following statements for each graph f(x), g(x) and h(x):

1. The given function is defined at x = 3.
2. The given function is continuous at x = 3.
3. The given function has removable discontinuity at x = 3.
4. The given function has jump discontinuity at x = 3.
5. The given function has infinite discontinuity at x = 3.
6. The limit of the given function, as x approaches 3, exists.
7. The given function is defined at x = -1.
8. The given function is continuous at x = -1.
9. The given function has removable discontinuity at x = -1.
10. The given function has jump discontinuity at x = -1.
11. The given function has infinite discontinuity at x = -1.
12. The limit of the given function, as x approaches -1, exists.

f(x):

1. The given function is defined at x = 3: TRUE. It is already shown in the graph.
2. The given function is continuous at x = 3. TRUE. Its left and right limit as x approaches 3 is equal to f(3).
3. The given function has removable discontinuity at x = 3. FALSE. It is already proven to be continuous at x = 3.
4. The given function has jump discontinuity at x = 3. FALSE. It is already proven to be continuous at x = 3.
5. The given function has infinite discontinuity at x = 3. FALSE. It is already proven to be continuous at x = 3.
6. The limit of the given function, as x approaches 3, exists. TRUE. Its left limit and right limit as x approaches 3 is the same.
7. The given function is defined at x = -1. FALSE. It is already shown in the graph.
8. The given function is continuous at x = -1. FALSE. Though its left and right limit as x approaches -1 is the same, the function itself is not defined at x = -1.
9. The given function has removable discontinuity at x = -1. TRUE. By the definition of removable discontinuity.
10. The given function has jump discontinuity at x = -1. FALSE. Its left and right limit as x approaches -1 are the same.
11. The given function has infinite discontinuity at x = -1. FALSE. It doesn't approach infinity or negative infinity.
12. The limit of the given function, as x approaches -1, exists. TRUE. Its left and right limit as x approaches -1 is equal, so the limit exists.

g(x):

1. The given function is defined at x = 3: TRUE. It is already shown in the graph.
2. The given function is continuous at x = 3. FALSE. Its left and right limit as x approaches 3 are NOT equal to f(3).
3. The given function has removable discontinuity at x = 3. TRUE. Its limit as x approaches 3 exists, and it is NOT equal to f(3).
4. The given function has jump discontinuity at x = 3. FALSE. Its left and right limit as x approaches 3 are the same.
5. The given function has infinite discontinuity at x = 3. FALSE. It doesn't approach infinity or negative infinity.
6. The limit of the given function, as x approaches 3, exists. TRUE. Its left limit and right limit as x approaches 3 is the same.
7. The given function is defined at x = -1. FALSE. It is already shown in the graph.
8. The given function is continuous at x = -1. FALSE. Left limit approaches negative infinity, right limit approaches infinity.
9. The given function has removable discontinuity at x = -1. FALSE. Since the left limit as x approaches -1, is not a finite value (it's negative infinity) and the right limit as x approaches -1, is also not a finite value (it's infinity), the limit as x approaches -1 doesn't exist.
10. The given function has jump discontinuity at x = -1. FALSE. Since the left limit as x approaches -1, is not a finite value (it's negative infinity) and the right limit as x approaches -1, is also not a finite value (it's infinity), it doesn't follow the definition of jump discontinuity.
11. The given function has infinite discontinuity at x = -1. TRUE. By the definition of infinite discontinuity.
12. The limit of the given function, as x approaches -1, exists. FALSE. Since the left limit as x approaches -1, is not a finite value (it's negative infinity) and the right limit as x approaches -1, is also not a finite value (it's infinity), the limit as x approaches -1 doesn't exist.

h(x):

1. The given function is defined at x = 3: FALSE. It is already shown in the graph.
2. The given function is continuous at x = 3. FALSE. f(3) is undefined.
3. The given function has removable discontinuity at x = 3. FALSE. Its left limit as x approaches 3 is NOT equal to its right limit, so the limit as x approaches 3 does not exist.
4. The given function has jump discontinuity at x = 3. TRUE. Both left and right limit as x approaches 3, are finite values. Its left limit as x approaches 3 is NOT equal to its right limit.
5. The given function has infinite discontinuity at x = 3. FALSE. It doesn't approach infinity or negative infinity.
6. The limit of the given function, as x approaches 3, exists. TRUE. Its left limit as x approaches 3 is NOT equal to its right limit, so the limit doesn't exist at x = 3.
7. The given function is defined at x = -1. TRUE. It is already shown in the graph.
8. The given function is continuous at x = -1. FALSE. Its left limit as x approaches -1 is NOT equal to its right limit, so the limit as x approaches -1 does not exist.
9. The given function has removable discontinuity at x = -1. FALSE. As proven before, the limit of the function as x approaches -1 does not exist. So, it can't be equal to f(-1).
10. The given function has jump discontinuity at x = -1. TRUE. By the definition of jump discontinuity.
11. The given function has infinite discontinuity at x = -1. FALSE. It doesn't approach infinity or negative infinity.
12. The limit of the given function, as x approaches -1, exists. FALSE. Its left limit as x approaches -1 is NOT equal to its right limit, so the limit as x approaches -1 does not exist.