How to Find the Limit of a Function Algebraically – 13 Best Methods

A Quick Summary of Limit

,A real number l is said to be the limit of a function f(x) at x=a if for every +ve \varepsilon there exists a +ve \delta such that

\left | f\left ( x \right ) - l \right |< \varepsilon, whenever 0 < \left | x - a \right |< \delta

and it is expressed as

\boldsymbol{\lim_{x \to {\color{Red} a}}{\color{Blue} f(x)}={\color{Green} l}}

Read more: Concept of Limit of a Function

How to find the limit of a function Algebraically

There are different ways to find the limit of a function algebraically.

In this article, we will know about the 13 best methods to find the limit of a function.

#1. Direct Substitution

In the substitution method we just simply plug in the value of x in the given function f(x) for the limit.

Look at the examples given below:

\lim_{x \to 3}5x=5\times {\color{Magenta} 3}=15

Here we just plugged in x=3.

Is not it easy??

Some other examples:

  • \lim _{x \to -1}\left ( 2+x \right )^{10}=\left( 2{\color{Magenta} -1} \right )^{10}=1^{10}=1
  • \lim _{x \to 1}\left ( x^{2}+x+3 \right )=({\color{Magenta} 1})^{2}+1+3=1+1+3=5
  • \lim _{x \to -5}\left ( 10 \right )=10
  • \lim _{x \to 3}\sqrt[3]{24+x}=\sqrt[3]{24+{\color{Magenta}3}}=\sqrt[3]{27}=3
  • \lim _{x \to -2}\frac{x^{3}-5x+3}{x^{2}+1}=\frac{({\color{Magenta}-2})^{3}-5({\color{Magenta}-2})+3}{({\color{Magenta}-2})^{2}+1}=\frac{-8+10+3}{4+1}=\frac{5}{5}=1

#2. Limits by Factoring

Evaluate \lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2} .

If we plugin x=2, we get

\lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2} =\frac{({\color{Magenta} 2})^{2}-5({\color{Magenta} 2})+6}{({\color{Magenta} 2})^{3}-5({\color{Magenta} 2})+2}=\frac{4-10+6}{8-10+2} =\frac{0}{0} , an indeterminate form

Hence the substitution method does not work to find the limit.

So we have to try another method to find the limit.

Now we try the Factor method.

\: \: \: \lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2}

=\lim _{x \to 2}\frac{\left ( x-3 \right ){\color{Magenta} \left ( x-2 \right )}}{{\color{Magenta} \left ( x-2 \right )}\left ( x^{2}+2x-1\right )}

=\lim _{x \to 2}\frac{\left ( x-3 \right )}{\left ( x^{2}+2x-1\right )}

=\frac{{\color{Magenta}2}-3}{{\color{Magenta} 2}^{2}+2({\color{Magenta} 2})-1} (by substituting x=2)



=-\frac{1}{7} , a finite number.

\therefore \lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2}=\frac{-1}{7}

#3 Common Denominator method

Evaluate \lim _{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x}

First, we try to find the limit by using the direct substitution method.

If we plug in x=0, we get

\lim _{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x}=\frac{\frac{1}{{\color{Magenta}0}+2}-\frac{1}{2}}{{\color{Magenta}0}}=\frac{\frac{1}{2}-\frac{1}{2}}{0}=\frac{0}{0}, an indeterminate form.

\therefore the direct substitution method did not work here.

Also, we can not factor the function \frac{1}{x+2}-\frac{1}{2}.

So we have to use any other method.

In this case, we use the Common Dinomenator method to find the limit.

\: \: \: \lim_{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x}

=\lim_{x \to 0}\frac{\frac{{\color{Magenta}2}\times 1}{{\color{Magenta}2}(x+2)}-\frac{1{\color{Magenta}(x+2)}}{2{\color{Magenta}(x+2)}}}{x}

=\lim_{x \to 0}\frac{\frac{2-x-2}{2(x+2)}}{x}

=\lim_{x \to 0}\frac{\frac{-x}{2(x+2)}}{x}

=\lim_{x \to 0}\frac{-{\color{Magenta}x}}{2(x+2)}\times \frac{1}{{\color{Magenta}x}}

=\lim_{x \to 0}\frac{-1}{2(x+2)}

=\frac{-1}{2({\color{Magenta}0}+2)} (by plugging in x=0)


#4. Expansion Method – Open up Parenthesis

Evaluate \lim _{x \to 0}\frac{(x+3)^{2}-9}{x}

If we plug in x=0, we get

\lim _{x \to 0}\frac{(x+3)^{2}-9}{x}=\frac{({\color{Magenta}0}+3)^{2}-9}{{\color{Magenta}0}}=\frac{9-9}{0}=\frac{0}{0}, an indeterminate form.

So direct substitution does not work here.

Also, see that here we can not apply the Factor method and Common Denominator method to find the limit.

Also, see that finding the limit of a function \frac{(x+3)^{2}-9}{x} using the Factor method and Common Denominator method is not possible.

Hence we use the 4th method i.e., Expansion method

\: \: \: \lim_{x \to 0}\frac{(x+3)^{2}-9}{x}

=\lim_{x \to 0}\frac{x^{2}+6x+9-9}{x}

=\lim_{x \to 0}\frac{x^{2}+6x}{x}

=\lim_{x \to 0}\frac{{\color{Magenta}x}(x+6)}{{\color{Magenta}x}}

=\lim_{x \to 0}(x+6)

={\color{Magenta}0}+6 (by plugging in x=0)


\therefore \lim _{x \to 0}\frac{(x+3)^{2}-9}{x}=6

#5. Limits by Rationalizing – Rationalization Method

Evaluate \lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}

If we plug in x=8, we get

\lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}=\frac{8-8}{\sqrt{8+1}-3}=\frac{0}{\sqrt{9}-3}=\frac{0}{3-3}=\frac{0}{0}, an indeterminate form

\therefore the Direct Substitution method has failed.

You can see that we can not factor \sqrt{x+1}-3. So the Factor method also fails.

Also, we can not use the Common Numerator method as there is no fraction in the numerator.

The Expansion Method also failed.

So what can we do to find the limit?

In such cases, we use the Rationalization Method to find the limit like this:

\: \: \: \lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}

=\lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}\times \frac{{\color{Magenta}\sqrt{x+1}+3}}{{\color{Magenta}\sqrt{x+1}+3}} (multiplying numerator and denominator by \sqrt{x+1}+3, the conjugate of of \sqrt{x+1}-3)

=\lim_{x \to 8}\frac{(x-8)(\sqrt{x+1}+3)}{(\sqrt{x+1})^{2}-(3)^{2}} (by using (a+b)(a-b)=a^{2}-b^{2})

=\lim_{x \to 8}\frac{(x-8)(\sqrt{x+1}+3)}{x+1-9}

=\lim_{x \to 8}\frac{{\color{Magenta}(x-8)}(\sqrt{x+1}+3)}{{\color{Magenta}x-8}}

=\lim_{x \to 8}\frac{\sqrt{x+1}+3}{1}

=\frac{\sqrt{{\color{Magenta} 8}+1}+3}{1} (by plugging in x=8)



=6, a finite number

\therefore \lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}=6

#6. Limit of Absolute Value function

Example: Find the limit if it exists

\lim_{x \to 3}\left (\frac{\left |x-3\right |}{x-3}\right ).


If we plug in x=2, we get

\lim_{x \to 3}\left (\frac{\left |x-3\right |}{x-3}\right )=\frac{\left |3-3\right |}{3-3}=\frac{0}{0}, an indeterminate form.

So we can’t find the limit by substitution.

We will find the limit by using the definition of Absolute Value function.

The Absolute Value function is defined by

Absolute value function

Using this definition we can write

Absolute value function mod x-3

Now we have the value of \frac{\left |x-3\right |}{x-3} when x\geq 0 and x<0

So we have to find both the right-hand limit and the left-hand limit.

\: \: \: \: \: \: \: \lim_{x \to 3+}\frac{\left |x-3\right |}{x-3}=\frac{x-3}{x-3}=1

and \lim_{x \to 3-}\frac{\left |x-3\right |}{x-3}=\frac{-(x-3)}{x-3}=-1

See that

\lim_{x \to 3+}\frac{\left |x-3\right |}{x-3}=1\neq -1=\lim_{x \to 3-}\frac{\left |x-3\right |}{x-3}

i.e., Right hand limit \neq Left hand limit

\therefore \lim_{x \to 3}\left (\frac{\left |x-3\right |}{x-3}\right ) does not exists.

#7. Limit of the form of sinx/x

Evaluate \lim_{x \to 0}\frac{\sin 3x}{x}

Substituting x=0, we get

\lim_{x \to 0}\frac{\sin 3x}{x}=\frac{\sin (3\times 0)}{0}=\frac{\sin 0}{0}=\frac{0}{0}, an indeterminate form.

Therefore we can not find the limit using Direct Substitution.

In such types of problems we use the Limit property given below:

Rule: \lim_{x \to 0}\frac{\sin x}{x}=1, where x is measured in radian (i.e., in circular measure).

Now first we transform \frac{\sin 3x}{x} in the form of \frac{\sin x}{x} then we find the limit like this:

\lim_{x \to 0}\frac{\sin 3x}{x}=\lim_{x \to 0}\frac{\sin 3x}{{\color{Magenta} 3}\times x}\times \frac{{\color{Magenta} 3}}{1}=3\lim_{x \to 0}\frac{\sin 3x}{3x}=3\times 1=3

Therefore \lim_{x \to 0}\frac{\sin 3x}{x}=3

Some other examples:

  • \lim_{x \to 0}\frac{\sin 4x}{8x}=\frac{1}{2}\lim_{x \to 0}\frac{\sin 4x}{4x}=\frac{1}{2}(1)=\frac{1}{2},
  • \lim_{x \to 0}\frac{x}{\sin x}=\frac{1}{\lim_{x \to 0}\frac{\sin x}{x}}=\frac{1}{1}=1,
  • \lim_{x \to 0}\frac{x-\sin x}{x}=\lim_{x \to 0}\left ( 1-\frac{\sin x}{x} \right )=1-1=0,
  • \lim_{x \to 0}\frac{\tan x}{x}=\lim_{x \to 0}\frac{\sin x}{x}\frac{1}{\cos x}=\lim_{x \to 0}\frac{\sin x}{x}\lim_{x \to 0}\frac{1}{\cos x}=1\times \frac{1}{\cos 0}=1(1)=1,

#8. Limit of the form of (e^x -1)/x

Rule: \lim_{x \to 0}\frac{e^{x}-1}{x}=1

Example 1: How to find the limit of a function \frac{e^{5h}-1}{3h} algebraically as x approaches zero.


\: \: \: \lim_{h \to 0}\frac{e^{5h}-1}{3h}

=\lim_{h \to 0}\frac{e^{5h}-1}{{\color{Magenta} 5}h}\frac{{\color{Magenta} 5}}{3}

=\frac{5}{3}\lim_{h \to 0}\frac{e^{5h}-1}{5h}

=\frac{5}{3}\times 1 (\because \lim_{x \to 0}\frac{e^{x}-1}{x}=1)


\therefore \lim_{h \to 0}\frac{e^{5h}-1}{3h}=\frac{5}{3}

Example 2: How do you find the limit of a function \frac{e^{px}-e^{qx}}{x} as x approaches 0.


\: \: \: \lim_{x \to 0}\frac{e^{px}-e^{qx}}{x}

=\lim_{x \to 0}\frac{(e^{px}-1)-(e^{qx}-1)}{x}

=\lim_{x \to 0}\frac{e^{px}-1}{x}-\lim_{x \to 0}\frac{e^{qx}-1}{x}

=\lim_{x \to 0}\frac{e^{px}-1}{{\color{Magenta} p}x}\frac{{\color{Magenta} p}}{1}-\lim_{x \to 0}\frac{e^{qx}-1}{{\color{Magenta} q}x}\frac{{\color{Magenta} q}}{1}

=1\left ( \frac{p}{1} \right )-1\left ( \frac{q}{1} \right ) (\because \lim_{x \to 0}\frac{e^{x}-1}{x}=1)


\therefore \lim_{x \to 0}\frac{e^{px}-e^{qx}}{x}=p-q

Example 3:

\: \: \: \lim_{x \to 0}\frac{e^{\log x}-1}{e^{x-1}-1}

=\lim_{x \to 0}\frac{x-1}{e^{x-1}-1} ( \because e^{\log x}=x )

=\lim_{z \to 0}\frac{z}{e^{z}-1} (where x-1=z; then z\rightarrow 0, whenx\rightarrow 1)

=\frac{1}{\lim_{z \to 0}\frac{e^{z}-1}{z}}

=\frac{1}{1} (\because \lim_{x \to 0}\frac{e^{x}-1}{x}=1)


\therefore \lim_{x \to 0}\frac{e^{\log x}-1}{e^{x-1}-1}=1

Also read: How to find the Limit of a Function using Squeeze Theorem

#9. Limit of the form of (a^x – 1)/x

Rule: \lim_{x \to 0}\frac{a^{x}-1}{x}=\log_{e}a,\:a>0

Example 1: \lim_{x \to 0}\frac{3^{x}-1}{x}=\log_{e}3 \: \left ( \because 3>0 \right )

Example 2: Show that \lim_{x \to 0}\frac{5^{x}-4^{x}}{x}=\log_{e}\left ( \frac{5}{4} \right )


\: \: \: \lim_{x \to 0}\frac{5^{x}-4^{x}}{x}

=\lim_{x \to 0}\frac{(5^{x}-1)-(4^{x}-1)}{x}

=\lim_{x \to 0}\frac{5^{x}-1}{x}-\lim_{x \to 0}\frac{4^{x}-1}{x}

=\log_{e}5-\log_{e}4 (\because 5,4>0 and \lim_{x \to 0}\frac{a^{x}-1}{x}=\log_{e}a, when a>0)

=\log_{e}\left ( \frac{5}{4} \right )

\therefore \lim_{x \to 0}\frac{5^{x}-4^{x}}{x}=\log_{e}\left ( \frac{5}{4} \right )

#10. Limit of the form of (x^n – a^n)/(x-a)

Rule: \lim_{x \to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}, where n is a rational number.

Example 1: Show that \lim_{x \to -2}\frac{x^{5}+32}{x+2}=80.


\: \: \: \lim_{x \to -2}\frac{x^{5}+32}{x+2}

=\lim_{x \to -2}\frac{x^{5}-(-2)^{5}}{x-(-2)}

=5\times (-2)^{5-1} (\because \lim_{x \to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1} and -2 is a rational number)

=5\times (-2)^{4}

=5\times 16


\therefore \lim_{x \to -2}\frac{x^{5}+32}{x+2}=80

Example 2: Prove that \lim_{h \to 0}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\frac{1}{3}x^{\frac{-2}{3}}.


\: \: \: \lim_{h \to 0}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}

=\lim_{h \to 0}\frac{(x+h)^{\frac{1}{3}}-x^{\frac{1}{3}}}{h}

If we put, x+h=z,then z\rightarrow x, when h\rightarrow 0

Therefore the given limit

=\lim_{z \to x}\frac{z^\frac{1}{3}-x^\frac{1}{3}}{z-x} (\because x+z=h,\: \therefore h=z-x)

=\frac{1}{3}x^{\frac{1}{3}-1} (\because \lim_{x \to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1})


\therefore \lim_{h \to 0}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\frac{1}{3}x^{\frac{-2}{3}}

#11. Limit of the form of (1/x)log(1+x)

Rule: \lim_{x \to 0}\frac{1}{x}\log_{e}\left ( 1+x \right )=1

Example: Show that \lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{x}=a


\: \: \: \lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{x}

=\lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{{\color{Magenta}a}x}\frac{{\color{Magenta}a}}{1}

=a\lim_{x \to 0}\frac{1}{ax}\log_{e}\left ( 1+ax \right )

=a\lim_{z \to 0}\frac{1}{z}\log_{e}\left ( 1+z \right ) ( where z=ax; then z\rightarrow 0, when x\rightarrow 0 )

=a (\because \lim_{x \to 0}\frac{1}{x}\log_{e}\left ( 1+x \right )=1)

\therefore \lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{x}=a

#12. Limit of the form of (1+x)^(1/x)

Rule: \lim_{x \to 0}\left ( 1+x \right )^{\frac{1}{x}}=e

Example 1: Prove that \lim_{x \to 0}\left ( 1+ax \right )^{bx}=e^{ab}.


\: \: \: \lim_{x \to 0}\left ( 1+ax \right )^{bx}

=\lim_{x \to 0}\left ( \left ( 1+ax \right )^\frac{1}{{\color{Magenta} ax}} \right )^{{\color{Magenta} ax}\times \frac{b}{x}}

=\left (\lim_{x \to 0} \left ( 1+ax \right )^\frac{1}{ax} \right )^{ab}

=\left (\lim_{z \to 0} \left ( 1+z \right )^\frac{1}{z} \right )^{ab} (where z=ax; then z\rightarrow 0, when x\rightarrow 0)

=(e)^{ab} (\because \lim_{x \to 0}\left ( 1+ax \right )^{bx}=e^{ab})


\therefore \lim_{x \to 0}\left ( 1+ax \right )^{bx}=e^{ab}

Example 2: Prove that \lim_{x \to 0}\left ( 1+4x \right )^{\frac{x+2}{x}}=e^{8}


\: \: \: \lim_{x \to 0}\left ( 1+4x \right )^{\frac{x+2}{x}}

=\lim_{x \to 0}(1+4x)^{1+\frac{2}{x}}

=\lim_{x \to 0}\left ( (1+4x)(1+4x)^{\frac{2}{x}} \right )

=\lim_{x \to 0}(1+4x)\times \lim_{x \to 0}(1+4x)^{\frac{1}{4x}\times 8}

=(1+4\times 0)\times \left ( \lim_{x \to 0}(1+4x)^{\frac{1}{4x}} \right )^{8}

=1\times \left ( \lim_{x \to 0}(1+4x)^{\frac{1}{4x}} \right )^{8}

=\left ( \lim_{x \to 0}(1+4x)^{\frac{1}{4x}} \right )^{8}

=\left ( \lim_{z \to 0}(1+z)^{\frac{1}{z}} \right )^{8} (where z=ax; then z\rightarrow 0, when x\rightarrow 0)

=(e)^{8} (\because \lim_{x \to 0}\left ( 1+x \right )^{\frac{1}{x}}=e)


\therefore \lim_{x \to 0}\left ( 1+4x \right )^{\frac{x+2}{x}}=e^{8}

#13. Limit of the form of ((1+x)^n-1)/x

Rule: \lim_{x \to 0}\frac{(1+x)^{n}-1}{x}=n

Example 1: Prove that \lim_{x \to 0}\frac{(1+x)^{6}-1}{2x}=3


\lim_{x \to 0}\frac{(1+x)^{6}-1}{2x}

=\lim_{x \to 0}\frac{(1+x)^{6}-1}{x}\left ( \frac{1}{{\color{Magenta} 2}} \right )

=\frac{1}{2}\times \lim_{x \to 0}\frac{(1+x)^{6}-1}{x}

=\frac{1}{2}\times 6 (\because \lim_{x \to 0}\frac{(1+x)^{n}-1}{x}=n)


\therefore \lim_{x \to 0}\frac{(1+x)^{6}-1}{2x}=3

We hope after reading this article you understand How to Find the Limit of a Function Algebraically.

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