Indeterminate form 0 x infinity

In this Post we will talk about One of Indeterminate forms i.e \(\;\;0 \times \infty\;\;\) so how to solve this?

Let's say we want to find \(\;\;\displaystyle \lim_{x \to c} (f(x) \times g(x))\;\;\), but when we put the limit ( or we approaches the limit) we get \(\;0 \times \infty \;\;\) form .

Sometimes students make mistake and just consider this \(\;0\;\), but reality is something different, To overcome such mistake use the following steps.

  1). First check whether it is a \(\; 0 \times \infty \;\) form or not , if 'yes' then follow step number 2 , if 'no' then check what kind of Indeterminate form it is. 

 2). Now do some rearrangements and make this a \(\;\displaystyle  \frac{0}{0} \; \) form or \(\;\displaystyle \frac{\infty}{\infty}\;\) form . generally if your question is in form \(\; f(x).g(x)\;\) form , then make it either in this \(\;\displaystyle \frac{f(x)}{1/g(x)}\;\) form or this \(\;\displaystyle \frac{g(x)}{1/f(x)}\;\) ( that depends upon question to question) 

3). Now you get a \(\; \displaystyle \frac{0}{0}\;\) form or \(\;\displaystyle \frac{\infty}{\infty}\;\) this is another Indeterminate form, and we already know that L'Hôpital's rule can easily deal with such kinds of Indeterminate forms.

 4). Keep on applying L'Hôpital's rule until you get rid of \(\; \displaystyle \frac{0}{0}\;\) form or \(\; \displaystyle \frac{\infty}{\infty}\;\). 

 5). At the end you will get your answer. 

 Let's try this with some examples

\(\displaystyle 1). \;\lim _{ x \rightarrow \displaystyle\frac{\pi}{2}}(1-\sin x) \tan x\)

SOLUTION: As \(x \rightarrow \displaystyle \frac{\pi}{2}\)

\(\displaystyle \Rightarrow (1-\sin x) \rightarrow 0 \; \text{And}\; \tan x \rightarrow \infty\)

so this is a \(\displaystyle \;0 \times \infty\;\)form

(NOTE : Sometimes we have to be careful about approaching limit form left or right, but in above question this will not going to change our answer)

Now as we discused in step number 2).

write given expression in \(\displaystyle\lim _{x \rightarrow \displaystyle \frac{\pi}{2}} \frac{(1-\sin x)}{1/ \tan x}\;\) form

Simplifying above we will get \(\displaystyle \;\;\lim _{x \rightarrow \displaystyle \frac{\pi}{2}} \frac{1-\sin x}{\cot x}\;\;\cdots (1)\)

\(\; \left[ \because \; \displaystyle \tan x=\frac{1}{\cot x} \right]\;\)

if you look carefully at \(\;(1)\;\) and take limit it is a \(\displaystyle \;\; \frac{0}{0}\;\;\) form

Because as\(\displaystyle \;\; x \rightarrow \displaystyle \frac{\pi}{2}\)

\(\displaystyle \Rightarrow (1-\sin x) \rightarrow 0 \;\; \text{And}\;\;\cot x \rightarrow 0\)

So \(\;(1)\;\) is a \(\displaystyle \;\frac{0}{0}\;\) Indeterminate form.

Now we know How to deal with \(\displaystyle \; \frac{0}{0} \;\) form, By using L'Hôpital's rule

See the diagram as \(\;x\;\) goes to \(\;\pi/2\;\) graph goes to \(\;0\).

By applying L'Hôpital's rule on \(\;(1)\;\) we get

\(\displaystyle \lim _{x \rightarrow \displaystyle \frac{\pi}{2}} \frac{\displaystyle \frac{d}{d x}(1-\sin x)}{\displaystyle \frac{d}{d x}(\cot x)}=\lim _{x \rightarrow \displaystyle \frac{\pi}{2}} \displaystyle \frac{-\cos x}{-\operatorname{cosec}^{2} x}\)

\(\displaystyle \Rightarrow \lim _{x \rightarrow \displaystyle \frac{\pi}{2}} \frac{\cos x}{\operatorname{cosec}^{2} x}\)

Now take limit \(\displaystyle x \rightarrow \frac{\pi}{2} \)

\(\displaystyle \Rightarrow \cos x \rightarrow 0 \;\;\;\;\; \text{And}\;\; \operatorname{cosec}^{2}x \rightarrow 1\)

so \(\displaystyle \frac{0}{1}=0\)


\(\displaystyle 2)\lim _{x \rightarrow \infty} 2^{x} \sin \left(\displaystyle \frac{a}{2^{x}}\right)\)
SOLUTION: As \( \;\;x \rightarrow \infty\)
\(\Rightarrow 2^{x} \rightarrow \infty \; , \;\sin \left(\displaystyle \frac{a}{2^{x}}\right) \rightarrow 0 \)
This is a \(\infty \times 0\) form we can write

\(\displaystyle \;\; \lim _{x \rightarrow \infty} 2^{x} \sin \left(\displaystyle \frac{a}{2^{x}}\right) \text { as } \)
\(\displaystyle \lim _{x \rightarrow \infty} \frac{\sin \left(\displaystyle \frac{a}{2^{x}}\right)}{\displaystyle \frac{1}{2^{x}}}\;\; \ldots \ldots \text { (1) }\)
As we take  \(\;\;x \;\rightarrow \infty\)
\(\displaystyle \Rightarrow \sin \left(\displaystyle \frac{a}{2^{x}}\right) \rightarrow 0, \displaystyle \frac{1}{2^{x}} \rightarrow 0\)

From above this is clear that \(\; (1)\;\) is a \(\;\frac{0}{0}\;\) Indeterminate form, so we will use L'Hôpital's rule Here
\(\displaystyle \Rightarrow \lim _{x \rightarrow \infty} \frac{\displaystyle \frac{d}{d x}\left[\sin \left(\frac{a}{2^{x}}\right)\right]}{\displaystyle \frac{d}{d x}\left(\displaystyle \frac{1}{2^{x}}\right)} \)

\(\displaystyle \lim _{x \rightarrow \infty} \frac{\cos \left(\displaystyle \frac{a}{2^{x}}\right) \times\left(-2^{-x} \ln 2\right) a}{-2^{-x} \ln 2} \quad\)

\(\left[\because \displaystyle \frac{d}{d x}\left(\displaystyle \frac{a}{2^{x}}\right)=-2^{-x} a\ln 2\right]\)
Cancel out terms, and we left with
\(\displaystyle \lim _{x \rightarrow \infty} \cos \left(\displaystyle \frac{a}{2 x}\right) \times a \)
\(\displaystyle \text{As} \;\;x \rightarrow \infty, \cos \left(\displaystyle \frac{a}{2^{x}}\right) \rightarrow 1 \)
\(\displaystyle \lim _{x \rightarrow \infty} \cos \left(\displaystyle \frac{a}{2^{x}}\right) \times a=1 \times a=a \)


Practice Questions 1). \(\displaystyle \lim_{x\to \infty} \; (2^{\displaystyle\frac{1}{x}} -1)x \)
ANSWER : \(\;\; \ln(2)\).
2). \(\displaystyle \lim_{x \to \displaystyle \frac{\pi}{2}} (\sec x)\ln\left(\displaystyle \frac{\pi}{2 x }\right) \)
ANSWER : \(\displaystyle \;\; \frac{2}{\pi}\)

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