Limit of a Strange Power

Calculus Level 1

lim ⁡ x → 0 + x x = ? \large \displaystyle\lim_{x\rightarrow 0^{+}}x^{x} = \, ?

This is a part of 10-seconds challenge .


The answer is 1.

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Yash Dev Lamba by Yash Dev Lamba , 23, India

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2 solutions

I still don't know how to do it in 10 seconds without knowing the graph xDD, but I did it this way (by using L'Hôpital's Rule): lim ⁡ x → 0 + x x = lim ⁡ x → 0 + e x ln ⁡ x \large\displaystyle\lim_{x\to 0^+}x^x=\displaystyle\lim_{x\to 0^+}e^{x\ln x} e ( lim ⁡ x → 0 + ln ⁡ x 1 x ) = e ( lim ⁡ x → 0 + 1 x − 1 x 2 ) \large e^{\left(\lim_{x\to 0^+}\dfrac{\ln x}{\frac{1}{x}}\right)}=e^{\left(\lim_{x\to 0^+}\dfrac{\frac{1}{x}}{-\frac{1}{x^2}}\right)} e ( lim ⁡ x → 0 + − x ) = e 0 = 1 \large e^{(\lim_{x\to 0^+}-x)}=e^0=1

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you must know that power (means strength) of x > l n x x>lnx i.e x x will dominate over l n x lnx

so if x → 0 + x\rightarrow0^{+} and l n x → − ∞ lnx \rightarrow -\infty then their product → 0 \rightarrow0

Yash Dev Lamba - 5 years, 3 months ago

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We can reach the same result considering the series expansion for ln after changing it into exponential form. And yes, with that the question indeed becomes a 10 sec challenge!

Pulkit Gupta - 5 years, 3 months ago

YEAH,VEGETA SSJ2 IS THE STRONGEST.

Manish Maharaj - 5 years, 1 month ago

@Hjalmar Orellana Soto : Hey there! Nice solution, thanks for posting it. +1! Keep up the good work.

But there were few formatting errors which I corrected. Just click toggle latex(in the menu on your profile pic) or click edit solution to see the changes made.

Try to use \lim instead of just typing "lim". Also, try using \displaystyle before the \lim expression to get a better view of it. For more information on latex, see the following guide: click here .

Sravanth C. - 5 years, 3 months ago

lim ⁡ x → 0 + x x x x = e x ln ⁡ ( x ) = e x ln ⁡ ( x ) Apply a limit chain rule = lim ⁡ x → 0 + e x ln ⁡ ( x ) g ( x ) = x ln ⁡ ( x ) , f ( u ) = e u = lim ⁡ x → 0 + e x ln ⁡ ( x ) ⇒ 0 Apply L’ Hopital’s rule = lim ⁡ x → 0 + e u = 1 \begin{aligned}\lim_{x\to 0^+} x^x \quad\quad\quad\quad\quad\quad{x^x = e^{x \ln(x)}} \\& = e^{x\ln(x)} \quad\quad\quad\quad\quad\quad{\text{Apply a limit chain rule}} \\& =\lim_{x \to 0^+} e^{x \ln(x)} \quad\quad\quad\quad\quad\quad{g(x) = x \ln(x), f(u) = e^u} \\& =\lim_{x\to 0^+} e^{x \ln(x)} \Rightarrow 0 \quad\quad\quad\quad\quad\quad{\text{Apply L' Hopital's rule}} \\& =\lim_{x \to 0^+} e^u \\& = 1 \end{aligned}

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