Question

lim(x----->0) {1/x}^x

Answer 1

I don't know of this helps, but: \[\lim_{x\rightarrow0}\left(\dfrac{1}{x}\right)^x=\lim_{x\rightarrow0}\dfrac{1^x}{x^x}=\dfrac{\lim_{x\rightarrow0}1^x}{\lim_{x\rightarrow0}x^x}=\dfrac{1}{\lim_{x\rightarrow0}x^x}\]

Answer 2

but 0^0 doesn't make any sense

Answer 3

\[\large\begin{align*}
\lim_{x\to0}\left(\frac{1}{x}\right)^x&=\lim_{x\to0^+}e^{\ln\left(\frac{1}{x}\right)^x}\\\\
&=\lim_{x\to0}e^{x\ln\frac{1}{x}}\\\\
&=e^{{\displaystyle\lim_{x\to0^+}}~x\ln\frac{1}{x}}\\\\
&=e^{{\displaystyle\lim_{x\to0^+}}~\dfrac{\ln\frac{1}{x}}{\frac{1}{x}}}\\\\
&=e^{{\displaystyle\lim_{x\to0^+}}~\dfrac{-\ln x}{\frac{1}{x}}}\\\\
&=e^{-{\displaystyle\lim_{x\to0^+}}~\dfrac{\ln x}{\frac{1}{x}}}\\\\
&=e^{-{\displaystyle\lim_{x\to0^+}}~\dfrac{\frac{1}{x}}{-\frac{1}{x^2}}}&\text{via L'Hopital's rule}\\\\
&=e^{{\displaystyle\lim_{x\to0^+}}~\dfrac{1}{\frac{1}{x}}}\\\\
&=e^{{\displaystyle\lim_{x\to0^+}}x}\\\\
&=e^0\\\\
&=1
\end{align*}\]

Answer 4

yeah it looks right

Answer 5

@micahwood50 kind of does help. I'm finish where he stopped.

Let's focus on limit only, assume it's one-sided.
\[\begin{align*}
\lim_{x\rightarrow0^+} x^x&= \lim_{x\rightarrow0^+} e^{\ln(x^x)} \\~\\
&=\lim_{x\rightarrow0^+}e^{x\ln(x)}\\~\\
&=e^{~\displaystyle\lim_{x\rightarrow0^+}x\ln (x)}~~~\text{This will get us form 0/}\infty,\\&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{ so let move x to denominator as 1/x}\\~\\
&=e^{~\displaystyle\lim_{x\rightarrow0^+}\frac{\ln (x)}{\frac{1}{x}}}~~~ \text{The form is now 0/0, so let's use L'Hospital's rule}\\~\\
&=e^{~\displaystyle\lim_{x\rightarrow0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}}\\~\\
&=e^{~\displaystyle\lim_{x\rightarrow0^+}\frac{-x^2}{x}}\\~\\
&=e^{~\displaystyle\lim_{x\rightarrow0^+}-x}\\~\\
&=e^0\\~\\
&=\boxed{1}
\end{align*}\]

So we have \(\dfrac{1}{\displaystyle \lim_{x\rightarrow0^+}x^x}=\dfrac{1}{1}=\boxed{1}\)

Answer 6

Though I am confused about how you can switch from two-sided limit to one sided limit... Can you explain? @SithsAndGiggles

Answer 7

@geerky42, the switch to a one-sided limit has to do with the introduction of the logarithm.

\(\ln\dfrac{1}{x}\) is only defined for \(\dfrac{1}{x}>0\), i.e. \(x>0\) \(\bigg(\)since all numbers \(\dfrac{1}{x}\) are positive, but \(\dfrac{1}{x}\) is undefined at \(x=0\bigg)\).