# Limits of Functions

The theory of limits of functions is the cornerstone of calculus because these are the limits upon which the notion of a derivative depends. What is a Limit? Let us understand the definition of “limit” with the help of an example, consider the function f(x). Let the independent variable x take values near a given constant ‘a’. Then, f(x) takes a corresponding set of values. Suppose that when x is close to ‘a’, the values of f(x) are close to some constant. Suppose f(x) can be made to differ arbitrarily small from A by taking values of x that are sufficiently close to ‘a’ but not equal to ‘a’ and that is true for all such values of x. Then, f(x) is said to approach limit A as x approaches ‘a’.

## Limits and Functions

A function may approach two different limits. One where the variable approaches its limit through values larger than the limit and the other where the variable approaches its limit through values smaller than the limit. In such a case, the limit is not defined but the right and left-hand limit exist.

The right-hand limit of a function is the value of the function approaches when the variable approaches its limit from the right.

This can be written as $\lim_{x\rightarrow a}$ f(x) = A+

The left-hand limit of a function is the value of the function approaches when the variable approaches its limit from the left.

This can be written as $\lim_{x\rightarrow a}$f(x) = A

 The limit of a function exists if and only if the left-hand limit is equal to the right-hand limit. $\lim_{x\rightarrow a^{-1}}$ f(x) = $\lim_{x\rightarrow a^{+}}$ f(x) = L

Note: The limit of this function exists between any two consecutive integers.

## Properties of Limits

If limits $\lim_{x \rightarrow a}$ f(x) and $\lim_{x \rightarrow a}$ g(x) exists, then,

 Law of Addition $\lim\limits_{x \to a} \left [ f(x) + g(x) \right ]= \lim\limits_{x \to a}f(x) + \lim\limits_{x \to a}g(x)$ Law of Subtraction $\lim\limits_{x \to a} \left [ f(x) – g(x) \right ]= \lim\limits_{x \to a}f(x) – \lim\limits_{x \to a}g(x)$ Law of Multiplication $\lim\limits_{x \to a} \left [ f(x) .g(x) \right ]= \lim\limits_{x \to a}f(x) . \lim\limits_{x \to a}g(x)$ Law of Division $\large \lim\limits_{x \to a} \left [ \frac{f(x)}{g(x)} \right ] = \frac{\lim\limits_{x \to a}f(x)}{\lim\limits_{x \to a}g(x)}$, where $\lim\limits_{x \to a}g(x) \neq 0$ Law of Constant $\lim\limits_{x \to a} c = c$ Law of Root $\lim_{x \rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \rightarrow a}f(x)}$ Law of Power $\lim\limits_{x \to a} (f(x)^n = (\lim\limits_{x \to a} f(x))^n$

Where, n is an integer.

 Special Rules: $\lim\limits_{x\to a}\frac{x^{n}-a^{n}}{x-a} =n\: a^{(n-1)}$ , for all real values of n. $\lim_{\theta \rightarrow 0}\frac{\sin \theta }{\theta }= 1$ $\lim_{\theta \rightarrow 0}\frac{\tan \theta }{\theta }= 1$ $\lim_{\theta \rightarrow 0}\frac{1-\cos \theta }{\theta } = 0$ $\lim_{\theta \rightarrow 0}\cos \theta = 1$ $\lim_{x\rightarrow 0} e^{x} = 1$ $\lim_{x\rightarrow 0} \frac{e^{x}-1}{x} = 1$ $\lim_{x\rightarrow \infty }(1+ \frac{1}{x} )^{x}= e$

## Limit of a Function of Two Variables

If we have a function f(x,y) which depends on two variables x and y. Then this given function has the limit say C as (x,y) → (a,b) provided that ϵ>0,∃ δ > 0 such that |f(x,y)−C| < ϵ whenever 0 < $\sqrt{(x-a)^2 + (y-b)^2}$ < δ

It is defined as $\lim_{(x,y) \rightarrow (a,b)}$f(x,y) = C

### Limits of Functions and Continuity

Limits and continuity are closely related to each other. Functions can be continuous or discontinuous. The continuity of a function is defined as, if there are small changes in the input of the function then must be small changes in the output.

In elementary calculus, the condition f(X) -> $\lambda$ as x -> a means that the number f(x) can be made to lie as close as we like to the number lambda as long we take the number x unequal to the number a but close enough to a. Which shows that f(a) might be very far from lambda and there is no need for f(a) even to be defined. The very important result we use for the derivation of function is: f'(a) of a given function f at a number a can be thought of as,

 f'(a) = $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$

### Limits of Complex Functions

To differentiate functions of a complex variable follow the below formula:

The function f(z) is said to be differentiable at z = z0 if

$\lim_{\Delta\ z \rightarrow 0} \frac{f(z_0 + \Delta\ z)-f(z_0)}{\Delta\ z}$ exists. Here ∆z = ∆x + i∆y

## Limits of Trigonometric Functions

There are few important limit properties that are involved in trigonometric functions. Let m be a real number in the domain of the given trig function.

1. $\lim_{x \rightarrow m}$ sin x = sin m

2. $\lim_{x \rightarrow m}$ tan x = tan m

3. $\lim_{x \rightarrow m}$ cos x = cos m

4.$\lim_{x \rightarrow m}$ sec x = sec m

5. $\lim_{x \rightarrow m}$ cosec x = cosec m

6. $\lim_{x \rightarrow m}$ cot x = cot m

 Two Special Trigonometric Limits: 1. $\lim_{x \rightarrow 0} \frac{sin\ x}{x}$ = 1 2. $\lim_{x \rightarrow 0} \frac{1- cos\ x}{x}$ = 0

## Limits of Exponential Functions

For any real number x, the exponential function f with the base a is f(x) = a^x where a>0 and a not equal to zero. Below are some of the important limits laws used while dealing with limits of exponential functions.

 For b > 1 $\lim_{x \rightarrow \infty}b^x = \infty$ $\lim_{x \rightarrow -\infty}b^x = 0$   For 0 < b < 1 $\lim_{x \rightarrow \infty}b^x = 0$ $\lim_{x \rightarrow -\infty}b^x = \infty$

## Limits of Functions Examples

Let us illustrate some of the examples to understand how to find limits of functions.

Example 1: Check for the limit, $\lim_{x \rightarrow 0}\ \frac{|sin\ x|}{x}$

Solution:

Since we have modulus function in the numerator, so let us evaluate right hand and left hand limits first.

RHL = $\lim_{h \rightarrow 0}\ \frac{|sin\ (h)|}{h}$ = 1

LHL = $\lim_{h \rightarrow 0}\ \frac{|sin\ (-h)|}{-h}$ = -1

As RHL and LHL revert different values, therefore, given limit does not exist.

Example 2: Evaluate $\lim_{x \rightarrow 0}\ \frac{x^3 cot\ x}{1 – cos\ x}$

Solution:

Simplify the function first:

(x3 cot x) / (1 – cos x) = (x3 cot x) / (1 – cos x) . (1 + cos x) / (1 + cos x)

= (x3 cos x)(1 + cos x) / sin x(1 – cos2 x)

= (x3 cos x)(1 + cos x) / sin3 x

Now,

$\lim_{x \rightarrow 0}\ \frac{(x^3 cos x)(1 + cos x)}{sin^3 x}$

= 2

[using $\lim_{x \rightarrow 0}\ \frac{sin^3 x}{ x^3}$ = 1]

Example 3: Find the limit. lim_(x-> 0) (tan x)/(sin x)

Solution: lim_(x-> 0) (tan x)/(sin x)

We know, tan x = sin x / cos x

= lim_(x-> 0) (1/cos x)

= 1

=> lim_(x-> 0) (tan x)/(sin x) = 1

Example 4: Evaluate lim_(x→0) sin(2x)

Solution:

lim_(x→0) sin(2x) = lim_(x→0) [sin(2x)/2×(2)]

= 2 lim_(x→0) sin(2x)/2

= 2

Example 5: Simplify lim_(x→0) (8x3 − 27) / (4x2 + 6x + 9)

Solution:

(8x3 − 27) can be written as (2x)3 − 33

Solve above expression using the formula, a3−b3 = (a−b)(a2+ab+b2)

So, (2x)3 − 33 = (2x – 3) (4x2 + 6x + 9)

Now,

lim_(x→0) (8x3 − 27) / (4x2 + 6x + 9) = lim_(x→0) (2x – 3) (4x2 + 6x + 9)/ (4x2 + 6x + 9)

= lim_(x→0) (2x−3)

Plugging x as 0, we get, 2(0)−3 = −3

So, f(x) approaches -3 as x approaches 0.

Example 6: If $f(x)=0$ is a quadratic equation such that $f(-\pi )=f(\pi )=0$ and $f\left( \frac{\pi }{2} \right)=-\frac{3{{\pi }^{2}}}{4}$, then what is $\underset{x\to -\pi }{\mathop{\lim }}\,\frac{f(x)}{\sin (sinx)}$?

Solution:

$f(x)={{x}^{2}}-{{\pi }^{2}} \\\underset{x\to -\pi }{\mathop{\lim }}\,\frac{{{x}^{2}}-{{\pi }^{2}}}{\sin (sinx)}\\=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(-\pi +h)}^{2}}-{{\pi }^{2}}}{\sin (sin(-\pi +h))} \\\underset{h\to 0}{\mathop{\lim }}\,\frac{-2h\pi +{{h}^{2}}}{-\sin (sin+h)} \\\underset{h\to 0}{\mathop{\lim }}\,\frac{h-2\pi }{\frac{-\sin (sinh)}{\sinh }\times \frac{\sinh }{h}}\\=2\pi$

Example 7: Find the value of $\underset{x\to 2}{\mathop{\lim }}\,\frac{{{2}^{x}}+{{2}^{3-x}}-6}{\sqrt{{{2}^{-x}}}-{{2}^{1-x}}}$.

Solution:

$\underset{x\to 2}{\mathop{\lim }}\,\frac{{{2}^{x}}+{{2}^{3-x}}-6}{\sqrt{{{2}^{-x}}}-{{2}^{1-x}}} \\=\underset{x\to 2}{\mathop{\lim }}\,\frac{{{({{2}^{x}})}^{2}}-6\times {{2}^{x}}+{{2}^{3}}}{\sqrt{{{2}^{x}}}-2}-2 \\ =\underset{x\to 2}{\mathop{\lim }}\,\frac{({{2}^{x}}-4)({{2}^{x}}-2)(\sqrt{{{2}^{x}}}+2)}{(\sqrt{{{2}^{x}}}-2)(\sqrt{{{2}^{x}}}+2)} \\=\underset{x\to 2}{\mathop{\lim }}\,\frac{({{2}^{x}}-4)({{2}^{x}}-2)(\sqrt{{{2}^{x}}}+2)}{({{2}^{x}}-4)} \\=\underset{x\to 2}{\mathop{\lim }}\,({{2}^{x}}-2)(\sqrt{{{2}^{x}}}+2)\\=({{2}^{2}}-2)(2+2)\\=8$

Example 8: Find the value of $\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}$.

Solution:

$\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} \\=\underset{x\to 2}{\mathop{\lim }}\,\frac{1+\sqrt{2+x}-3}{\left( \sqrt{1+\sqrt{2+x}}+\sqrt{3} \right)(x-2)} \\=\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{2+x}-2}{\left( \sqrt{1+\sqrt{2+x}}+\sqrt{3} \right)(x-2)} \\=\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)}{\left( \sqrt{1+\sqrt{2+x}}+\sqrt{3} \right)\left( \sqrt{2+x}+2 \right)(x-2)} \\=\frac{1}{(2\sqrt{3})4}\\=\frac{1}{8\sqrt{3}}$