Limiting Problem

Calculus Level 1

What is the limit of the function as "x" approaches infinity ?


The answer is 1.

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2 solutions

Abyoso Hapsoro
Apr 11, 2015

lim x 1 x 1 x = 1 1 = 1 0 = 1 1 = 1 \lim _{ x\rightarrow \infty }{ \frac { 1 }{ { x }^{ \frac { 1 }{ x } } } } =\frac { 1 }{ { \infty }^{ \frac { 1 }{ \infty } } } =\frac { 1 }{ { \infty }^{ 0 } } =\frac { 1 }{ 1 } =1

I n c o m p l e t e \hookrightarrow Incomplete S o l u t i o n Solution
How did you directly evaluate 0 \boxed{\infty^{0}} as 0 \infty^0 is an i n d e t e r m i n a t e indeterminate f o r m form

Akhil Bansal - 5 years, 9 months ago

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I don't believe 0 \infty^{0} is indeterminant.

Jerry McKenzie - 3 years, 5 months ago
L N
Aug 2, 2014

Observe that this is equivalent to lim x 1 lim x x x \frac{\lim{x \to \infty}1}{\lim{x \to \infty} \sqrt[x]{x}} . We know that x x = x 1 / x = e ln x x \sqrt[x]{x} = x^{1/x} = e^{\frac{\ln{x}}{x}} . Since, ln x \ln{x} grows slower than x x , x x = 1 \sqrt[x]{x} = 1 as x goes to infinity. Hence, the original limit goes to 1 as well.

Another possible way to evaluate this is rewriting the limit using the substitution m = 1 x m=\dfrac{1}{x} as, lim m 0 ( m m ) \displaystyle \lim_{m\to 0}\bigg(m^m\bigg) If you evaluate it now, the value comes out as 0 0 0^0 which is taken as 1 1 most of the time but that isn't true for all cases and the value is undefined! Checking the limit by substituting values of x x like x = 0.00001 x=0.00001 gives us a result tending to 1 1 . So, the value of the limit is 1 1 . This concludes the solution.

Prasun Biswas - 6 years, 5 months ago

This looks to be the most comprehensive one.

Sombir Palit - 5 years, 11 months ago

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