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
\(\displaystyle \lim_{x \to c} f(x)=\infty\;\)
and
\(\displaystyle\lim_{x \to c}g(x)=\infty\;\),
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) \(\displaystyle \lim _{x \rightarrow \infty} \displaystyle\frac{\log x}{x^{n}}, n>0\)
Solution:
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'
\(\displaystyle \text { i.e } \lim _{x \rightarrow \infty}\displaystyle \frac{1}{x^{n}} \longrightarrow 0\)
\(\Rightarrow \displaystyle \lim_{x \to \infty} \frac{\log x}{x^n}\to 0\)
ANSWER: O
2) \(\displaystyle \lim _{x \rightarrow \infty} \frac{x^{4}}{e^{x}}\)
Solution:
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 \(\displaystyle \lim _{x \rightarrow \infty} \frac{24}{e^{x}} \longrightarrow 0\)
\(\displaystyle \Rightarrow \lim _{x \rightarrow \infty} \frac{x^{4}}{e^{x}}=0\)
ANSWER : 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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