How to find the Limit of a Function? - Techniques & Examples

Introduction

The idea of limit is the basis of all calculus concepts. We can find the limit of a function by different techniques. In this article, you will learn all four different techniques of solving limits.

4 ways of solving Limit of a Function

The limit of a function can be evaluated by four methods i.e. by substituting the value of x, factorizing, rationalizing the numerator and finding the lowest common denominator.

To solve limit of a function f(x) = L There are following steps that you can use.

1. Plug the value of x in the function to find the value of limit.
2. If the function is in the form of a quadratic equation, use the factorization method to simplify it then plug the value of x in it.
3. If the function is in the form of fraction, use the rationalisation method or find the lowest common denominator.
4. If the function is exponential, use the rules of limit to solve it.

Here are examples to solve the limit of a function using different methods.

Finding limit by plugging the value of x

It is a first and easy method to solve limits algebraically. This method does not work when the result becomes undefined by plugging the value of x. This method works only when the function is continuous.

How do you plug in limits?

Suppose we have to find the limit of the following function by plugging method.

$$\lim_{x→2}(x^3\;-\;3x^2\;+\;6x\;-\;3)$$

To calculate limit, we will use the following steps,

$$\lim_{x→2}(x^3\;-\;3x^2\;+\;6x\;-\;3\;=\;\lim_{x→2}(x^3)\;-\;\lim_{x→2}(3x^2)\;+\;\lim_{x→2}(6x)\;-\;\lim_{x→2}(3)$$ $$\lim_{x→2}(x^3\;-\;3x^2\;+\;6x\;-\;3)\;=\;(2^3)\;-\;3(2^2)\;+\;6(2)\;-\;3$$ $$\lim_{x→2}(x^3\;-\;3x^2\;+\;6x\;-\;3)\;=\;8\;-\;6(4)\;+\;12\;-\;3$$ $$\lim_{x→2}(x^3\;-\;3x^2\;+\;6x\;-\;3)\;=\;8\;-\;24\;+\;12\;-\;3$$ $$\lim_{x→2}(x^3\;-\;3x^2\;+\;6x\;-\;3)\;=\;17\;-\;12$$ $$\lim_{x→2}(x^3\;-\;3x^2\;+\;6x\;-\;3)\;=\;5$$

Finding the limit by Factorization

This method is valid for a function in which the plugging method fails. In other words, if the function is not continuous or becomes undefined by plugging the value of x. Factorization is a technique of soving an algebraic expression by making factors. So, we can convert any function in a simple form by using this technique.

Limits by Factoring example

Suppose we have to find the limit of a continuous function,

$$\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}$$

In this method, the limit of a function is calculated by converting it in the form of factors.

$$\;\lim_{x→5}\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}$$

Now, we can the about function as,

$$\;\lim_{x→5}\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}\;=\;\lim_{x→5}\frac{x^2\;-\;4x\;-\;2x\;+\;8}{x\;-\;4} $$ $$\;\lim_{x→5}\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}\;=\;\lim_{x→5}\frac{x(x\;-\;4)\;-2(x-4)}{x\;-\;4} $$ $$\;\lim_{x→5}\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}\;=\;\lim_{x→5}\frac{(x\;-\;4)(x\;-\;2)}{x\;-\;4} $$

Vanishing the similar terms,

$$\;\lim_{x→5}\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}\;=\;\lim_{x→5}(x\;-\;2) $$

Now, substitute the value of x in the function,

$$\;\lim_{x→5}\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}\;=\;(5\;-\;2)\;=\;3 $$

Hence,

$$\;\lim_{x→5}\frac{x^2\;-\;6x\;+\;8}{x\;-\;4}\;=\;03 $$

Finding Limit by rationalization

In this limit solving technique, we rationalise the function by multiplying it with its conjugate. The conjugate of a term is obtained by changing the sign. It is also valid for those cases where the plugging technique fails. Here is an example relation to this technique.

How do you find limits of rationalization?

Suppose we have to find the limit of the following function.

$$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x\;-\;13}$$

To rationalise it, multiply and divide by the conjugate of numerator.

$$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x\;-\;13}\;=\;\lim_{x→13}\left[\frac{\sqrt{x\;-\;4}-3}{x-13}\;\times\;\frac{\sqrt{x-4}+3}{\sqrt{x-4}+3}\right]$$

Simplifying,

$$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x-13}\;=\;\lim_{x→13}\left[\frac{(x\;-\;4)^2\;-\;3^2}{(x\;-\;13)(\sqrt{x\;-\;4}\;+\;3)}\right]$$ $$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x-13}\;=\;\lim_{x→13}\left[\frac{x\;-\;4\;-9}{(x\;-\;13)(\sqrt{x\;-\;4}\;+\;3)}\right]$$ $$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x-13}\;=\;\lim_{x→13}\left[\frac{x\;-\;13}{(x\;-\;13)(\sqrt{x\;-\;4}\;+\;3)}\right]$$

Vanishing the same terms, we get,

$$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x-13}\;=\;\lim_{x→13}\left[\frac{1}{(\sqrt{x\;-\;4}\;+\;3)}\right]$$

Now, we can easily plug the value of x in it.

$$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x-13}\;=\;\left[\frac{1}{\sqrt{13-4}+3}\right]$$ $$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x-13}\;=\;\left[\frac{1}{(\sqrt{9}\;+\;3)}\right]\;=\;\left[\frac{1}{(\sqrt{3}\;+\;3)}\right]$$ $$\lim_{x→13}\frac{\sqrt{x\;-\;4}\;-\;3}{x-13}\;=\;\frac{1}{6}$$

Finding the Lowest Common Denominator

It is another limit solving technique that is useful where the plugging technique fails. This is valid for those cases where the function is neither factorable nor can be rationalised.

How to find limit of a fraction by LCD?

Suppose we have to find the limit of following function, for which above three techniques fail.

$$\lim_{x→0}\frac{\frac{1}{x\;+\;6}\;-\;\frac{1}{6}}{x}$$

This function cannot be rationalised or factorised. So, here we will find the lowest common denominator to solve the limit.

$$\lim_{x→0}\frac{\frac{1}{x\;+\;6}\;-\;\frac{1}{6}}{x}\;=\lim_{x→0}\;\frac{\frac{6\;-\;(x\;+\;6)}{(x\;+\;6)6}}{x} $$ $$\lim_{x→0}\frac{\frac{1}{x\;+\;6}\;-\;\frac{1}{6}}{x}\;=\lim_{x→0}\;\frac{\frac{6\;-\;x\;-\;6}{(x\;+\;6)6}}{x}$$ $$\lim_{x→0}\frac{\frac{1}{x\;+\;6}\;-\;\frac{1}{6}}{x}\;=\lim_{x→0}\;\frac{-x}{6x(x\;+\;6)} $$ $$\lim_{x→0}\frac{\frac{1}{x\;+\;6}\;-\;\frac{1}{6}}{x}\;=\lim_{x→0}\;\frac{-1}{6(x\;+\;6)} $$ Now, using the limit of x in it. $$\lim_{x→0}\frac{\frac{1}{x\;+\;6}\;-\;\frac{1}{6}}{x}\;=\;\frac{-1}{6(0\;+\;6)}\;=\;\frac{-1}{6(6)}$$ $$\lim_{x→0}\frac{\frac{1}{x\;+\;6}\;-\;\frac{1}{6}}{x}\;=\;\frac{-1}{36}$$

How to find limit when x approaches infinity?

If the limit of a function is of the form,

$$f(x)\;=\;L$$

We say that f has a limit L when x approaches infinity. This kind of limit is called the infinite limit. Let’s discuss in an example to understand how to solve infinite limits.

Limit at infinity example

$$\text{Evaluate}\;\;\left[5\;+\;\frac{3}{x^2}\right] $$

This limit is an infinite limit. We can solve it by substituting the value of limit.

$$\left[5\;+\;\frac{3}{x^2}\right]\;=\;\left[5+\frac{3}{∞^2}\right]$$ $$\frac{1}{∞}\;=\;0$$

So,

$$\left[5\;+\;\frac{3}{x^2}\right]\;=\;(5+0)\;=\;5$$

FAQ’s

How to find the Limit of a Function as x approaches 0?

If we simply put the value of limit in the function, the limit will become undefined. So, we can use other techniques of solving limits. Consider an example, where we have to evaluate

$$\frac{2(-3\;+\;x)^2\;-\;18}{x} $$

If we put zero in the function, the limit will be undefined. So, using factoring technique,

$$\frac{2(-3\;+\;x)^2\;-\;18}{x}\;=\;\frac{2(x^2\;-\;6x\;+\;9)\;-\;18}{x} $$ $$\frac{2(-3\;+\;x)^2\;-\;18}{x}\;=\;\frac{2x^2\;-\;12x\;+\;18\;-\;18}{x} $$ $$\frac{2(-3\;+\;x)^2\;-\;18}{x}\;=\;\frac{2x^2\;-\;12x}{x} $$ $$\frac{2(-3\;+\;x)^2\;-\;18}{x}\;=\;\frac{x(2x\;-\;12)}{x} $$ $$\frac{2(-3\;+\;x)^2\;-\;18}{x}\;=\;\lim_{x→0}(2x\;-\;12)$$ Now, after simplification we can easily substitute the value of x. $$\frac{2(-3\;+\;x)^2\;-\;18}{x}\;=\;-12$$

How to find the Limit of a Function as x approaches a Number?

When x approaches a number, we can easily evaluate the limit by using four techniques. I.e. by plugging the value of x, factorising, rationalising or by finding the lowest numerator.

Let’s discuss in an example.

Example

Find the real limit of the given function if it exists.

$$\frac{x^2\;-\;3x\;+\;2}{x\;-\;1}$$ As the first technique fails for this limit, now we will use another technique, $$\frac{x^2\;-\;3x\;+\;2}{x\;-\;1}\;=\;\frac{x^2\;-\;2x\;-\;x\;+\;2}{x\;-\;1} $$ $$\frac{x^2\;-\;3x\;+\;2}{x\;-\;1}\;=\;\frac{x(x\;-\;2)\;-1(x\;-\;2)}{x\;-\;1} $$ $$\frac{x^2\;-\;3x\;+\;2}{x\;-\;1}\;=\;\frac{(x\;-\;2)(x\;-\;1)}{x\;-\;1} $$ $$\frac{x^2\;-\;3x\;+\;2}{x\;-\;1}\;=\;(x\;-\;2) $$

Now using the value of x,

$$\frac{x^2\;-\;3x\;+\;2}{x\;-\;1}\;=\;1\;-\;2\;=\;-1$$

What is the need of Limits in Mathematics?

Limits are essential in calculus and mathematical analysis. They are used to define continuity, derivatives and integrals. Moreover, the generalized form of limits are used in topological space. Limits express the concepts of infinite small and infinite large quantities in mathematics.