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.
The value of \( \lim _{x \rightarrow \infty}(\log x)^{1 / x} \)
Limit
L'hospital Rule
But try the problem first...
Answer: is (C)
ISI MStat 2018 PSA Problem 8
Introduction to Real Analysis by Bertle Sherbert
First hint
What is the form of the limit? Can you convert it to some known limit?
\(\infty^0\)
\(\frac{\infty}{\infty}\)
L'Hospital Rule
Second Hint
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 .
Final Step
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 .
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.
The value of \( \lim _{x \rightarrow \infty}(\log x)^{1 / x} \)
Limit
L'hospital Rule
But try the problem first...
Answer: is (C)
ISI MStat 2018 PSA Problem 8
Introduction to Real Analysis by Bertle Sherbert
First hint
What is the form of the limit? Can you convert it to some known limit?
\(\infty^0\)
\(\frac{\infty}{\infty}\)
L'Hospital Rule
Second Hint
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 .
Final Step
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 .
