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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

Key Concepts


Limit

L'hospital Rule

Check the Answer


Answer: is (C)

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 .

Similar Problems and Solutions



ISI MStat 2018 PSA Problem 8
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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

Check the Answer


Answer: is (C)

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 .

Similar Problems and Solutions



ISI MStat 2018 PSA Problem 8
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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