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

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

Limit

L'hospital Rule

## Check the Answer

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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