<s>[INST] How to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$ I tried like this:
Let $y=a^{2x}-2\Rightarrow a^{2x}=y+2\Rightarrow 2x\ln a=\ln\left(y+2\right)\Rightarrow x=\dfrac{\ln\left(y+2\right)}{2\ln a}$
Also if $x\longrightarrow0,$ then $y\longrightarrow a^{2(0)}-2=-1.$
But we I put each and every this assumption in the given expression, then I get hanged due to $x^x.$ How to use algebra or any other easy procedure to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$.
 [/INST] The secret here is to express $\dfrac{a^{2x}-2}{x^x}$ a function of the $\dfrac{e^{y}-1}{y}$, with $y=(2\cdot \ln a)\cdot x$, and function the $x\cdot \ln x$.
Note that
\begin{align}
\dfrac{a^{2x}-2}{x^x}
=&
\dfrac{a^{2x}-1}{x^x}-\dfrac{1}{x^x}
\\
=&
\dfrac{(e^{\ln a})^{2x}-1}{2x}\cdot\dfrac{2x}{x^x}-\dfrac{1}{x^x}
\\
=&
\dfrac{e^{2(\ln a)x}-1}{2x}
\cdot 
\dfrac{2x}{e^{x\ln x}}
-
\dfrac{1}{e^{x\ln x}}
\\
=&
\left(
(\ln a)
\cdot\dfrac{e^{2(\ln a)x}-1}{2(\ln a)x}
\cdot 
2x
-
1
\right)
\cdot 
\frac{1}{e^{x\cdot \ln x}}
\\
\end{align}
We have

*

*$\displaystyle\lim_{x\to 0}\dfrac{e^{2(\ln a)x}-1}{2(\ln a)x}=1$ implies
$\displaystyle\lim_{x\to 0}(\ln a)
\cdot\dfrac{e^{2(\ln a)x}-1}{2(\ln a)x}
\cdot 
2x=0$


*$\displaystyle\lim_{x\to 0}x\cdot \ln x=\displaystyle\lim_{x\to 0}\dfrac{\ln x}{\left(\dfrac{1}{x} \right)}=\displaystyle\lim_{x\to 0}\dfrac{D(\ln x)}{D\left(\dfrac{1}{x} \right)}=\displaystyle\lim_{x\to 0}\dfrac{\left(\frac{1}{x} \right)}{\left(-\frac{1}{x^2}\right)}=\displaystyle\lim_{x\to 0}\dfrac{1}{-\left(\frac{1}{x}\right)}=\displaystyle\lim_{x\to 0}-x=0$
implies $\displaystyle\lim_{x\to 0}\dfrac{1}{e^{x\cdot \ln x}}=\dfrac{1}{e^{(\; \lim_{x\to 0}x\cdot \ln x)}}=\dfrac{1}{e^{0}}=1$
</s>