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# ISI MStat 2018 PSA Problem 8 | Limit of a Function

This is a beautiful problem from ISI MSAT 2018 PSA problem 8 based on limit . We provide sequential hints so that you can try .

This is a beautiful problem from ISI MStat 2018 PSA Problem 8 based on limit of a function. Try yourself and use hints if required.

## Limit of a Function – ISI MStat Year 2018 PSA Question 8

The value of $\lim _{x \rightarrow \infty}(\log x)^{1 / x}$

• (A) is e
• (B) is 0
• (C) is 1
• (D) does not exist

### Key Concepts

Limit

L’hospital Rule

ISI MStat 2018 PSA Problem 8

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

What is the form of the limit? Can you convert it to some known limit?

$\infty^0$
$\frac{\infty}{\infty}$
L’Hospital Rule

Let , $\lim _{x \rightarrow \infty}(\log x)^{1 / x}$ =l (say) then taking log on both sides we get , $\lim _{x \rightarrow \infty} \frac{ log( logx) }{x}$ =log (l) . Now we will apply L’hospital rule .

Applying L’hospital rule we get , log (l)= $\lim _{x \rightarrow \infty} \frac{1}{logx x}$= 0 $\Rightarrow l=e^0 \Rightarrow l=1$

Hence , option (C) is correct .

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