Problem 2

Calculus Level 1

$Evaluate\quad \lim _{ x\rightarrow \infty }{ \quad \sqrt [ x ]{ x } }$

The answer is 1.

3 solutions

Kenny Lau
Jul 11, 2014

$\Large\begin{array}{rcl} \lim_{x\rightarrow\infty}\sqrt[x]x&=& e^{\lim_{x\rightarrow\infty}\frac{\ln x}x}\\ &=& e^{\lim_{x\rightarrow\infty}\frac1x}\\ &=& e^0\\ &=&1 \end{array}$

Note : This solution is exactly the same as that of @Ronak Agarwal .

George Chen
Aug 6, 2014

Plugging it in, we have infinity to the infinity-th root, which can be rewritten as ∞ ^ (1/∞), which is ∞ ^ 0, which is 1.

Kïñshük Sïñgh
Jul 9, 2014

This can be written as x^1/x.. Thn on applying limit...we get something ( tending to infinity) ^0...that's equal to 1

Your solution is theorotically incorrect. It should be like this: $We\quad have\quad \underset { x\rightarrow \infty }{ lim } { x }^{ \frac { 1 }{ x } }=\underset { x\rightarrow \infty }{ lim } e^{ \frac { ln(x) }{ x } }\quad Applying\quad L-hopital's\\ rule\quad we\quad get\quad \underset { x\rightarrow \infty }{ lim } e^{ \frac { (\frac { 1 }{ x } ) }{ 1 } }=1$

Ronak Agarwal - 6 years, 11 months ago

But L-hopital's rule can only be applied when there is something in numerator as well as in denominator... And that too on applying x=a... Should be of 0/0 indeterminate form...

Kïñshük Sïñgh - 6 years, 11 months ago

It can also be applied when their is form of infinity/infinity

Ronak Agarwal - 6 years, 11 months ago

Problem 2

The answer is 1.

3 solutions

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