Solving Indeterminate form infinity/infinity

In this post we are going to solve INDETERMINATE form \(\displaystyle \;\; \frac{\infty}{\infty}\). 

We want to solve \(\displaystyle \lim_{x \to c}\frac{f(x)}{g(x)}\).

 But when you fill the limit you get 

\(\displaystyle \lim_{x \to c} f(x)=\infty\;\)


\(\displaystyle\lim_{x \to c}g(x)=\infty\;\),

then above form will bacomes a  \(\;\frac{\infty}{\infty}\;\). form.

 Which is one of  INDETERMINATE forms, to solve this we make use L'Hôpital's rule, How to use  L'Hôpital's rule we have discussed in this post.

1) \(\displaystyle \lim _{x \rightarrow \infty} \displaystyle\frac{\log x}{x^{n}}, n>0\)


as \(\displaystyle x \rightarrow \infty \;\Rightarrow\;\log x \rightarrow \infty\)

and given that  n>0.

as \(\displaystyle x \rightarrow \infty \;\Rightarrow\; x^{n} \longrightarrow \infty\)

So if we take limit then wo will get \(\displaystyle \frac{\infty}{\infty}\) form 

Now we will use  L'Hôpital's rule Here \( \displaystyle \lim _{x \rightarrow \infty} \frac{ \displaystyle \frac{d}{d x}(\log x)}{\displaystyle \frac{d}{d x} x^{n}}\)

\(\displaystyle \Rightarrow \lim _{x \rightarrow \infty} \displaystyle \frac{\displaystyle \frac{1}{x}}{n x^{n-1}} \)

From exponential formulas we know that 

 \( \displaystyle x^{1+n-1}=x^{n}\)

\(\displaystyle\Rightarrow \lim _{x \rightarrow \infty} \displaystyle \frac{1}{n x^{n}}\cdots \;\;(1)  \)

 Now as we have n>0

So as \(\; \displaystyle x \rightarrow \infty \quad x^{n} \longrightarrow \infty\)

and we know that when denominator goes to infinity and numerator is some finite number then whole thing will goes to '0'

 \(\displaystyle \text { i.e } \lim _{x \rightarrow \infty}\displaystyle \frac{1}{x^{n}} \longrightarrow 0\)

\(\Rightarrow \displaystyle \lim_{x \to \infty} \frac{\log x}{x^n}\to 0\)


2) \(\displaystyle \lim _{x \rightarrow \infty} \frac{x^{4}}{e^{x}}\)


as \(\displaystyle x \rightarrow \infty, x^{4} \longrightarrow \infty\)

and \(\displaystyle x \rightarrow \infty, e^{x} \longrightarrow \infty\)

so \(\;\displaystyle \lim _{x \rightarrow \infty} \frac{x^{4}}{e^{x}}\; \) is a 

\(\;\displaystyle \frac{\infty}{\infty}\; \) Form 

So we will use  L'Hôpital's rule

Take \(\displaystyle \lim _{x \rightarrow \infty} \frac{\frac{d}{d x}\left(x^{4}\right)}{\frac{d}{d x}\left(e^{x}\right)}=\frac{4 x^{3}}{e^{x}}\)

Now take \(\displaystyle \lim _{x \rightarrow \infty} \frac{4 x^{3}}{e^{x}}\)

Which is again \(\; \displaystyle \frac{\infty}{\infty}\; \)

Again applying   L'Hôpital's rule we will get

\(\displaystyle \lim _{x \rightarrow \infty} \frac{12 x^{2}}{e^{x}}\)

which is again \(\; \displaystyle \frac{\infty}{\infty}\; \) form

 again apply  L'Hôpital's rule we will get

\(\displaystyle \lim _{n \rightarrow \infty} \displaystyle \frac{\displaystyle \frac{d}{d x}\left(12 x^{2}\right)}{\displaystyle \frac{d}{d x}\left(e^{x}\right)}\)

\(\displaystyle \lim _{x \rightarrow \infty} \frac{24 x}{e^{x}}\)

Take limit we will get \(\;\displaystyle \frac{\infty}{\infty}\;\;\)form. 

one more time take  L'Hôpital's rule

\(\displaystyle \lim _{x \rightarrow \infty} \frac{\displaystyle\frac{d}{d x}(24 x)}{\displaystyle \frac{d}{d x}\left(e^{x}\right)}\)

\(\displaystyle \Rightarrow \lim _{x \rightarrow \infty} \displaystyle \frac{24}{e^{x}}\cdots\;\;\) (II)

 Now if you take limit of (II) it will Not be a INDETERMINATE form . So you have seen in above Example we have to apply  L'Hôpital's rule Rule 4 times, so this depends upon question to question ,keep applying  L'Hôpital's rule until you not get rid of \(\;\displaystyle \frac{\infty}{\infty}\;\)form.

Because \(\displaystyle x \rightarrow \infty \Rightarrow e^{x} \rightarrow \infty\)

So now above \(\displaystyle \lim _{x \rightarrow \infty} \frac{24}{e^{x}} \rightarrow 0\)

so \(\displaystyle \lim _{x \rightarrow \infty} \frac{24}{e^{x}} \longrightarrow 0\)

\(\displaystyle \Rightarrow \lim _{x \rightarrow \infty} \frac{x^{4}}{e^{x}}=0\)


Try to solve these Exercise  questions :

1). \(\displaystyle \lim_{x \to \infty} \frac{x^n}{e^x}\)

ANSWER : \(n!\)

2). \(\displaystyle \lim_{x \to 1} \frac{\log(1-x)}{x^2}\)

ANSWER : \(1\)

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