<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] Given
\begin{equation}
\lim_{x\rightarrow 0}\dfrac{a^{2x}-2}{x^x}=-1
\implies
\lim_{x\rightarrow 0}\dfrac{a^{2x}-2}{x^x} +1 =0
\end{equation}
You say,"then I get hanged due to  $^$.." but $0^0=1$ and we can even eliminate the denominator if that isn't enough.
\begin{align*}
(\large{x}^x)
\small{\bigg(\dfrac{a^{2x}-2}{x^x}} + 1\bigg)=0
\implies 
\space&a^{2 x} + x^x - 2=0\\ \\
 a^{2 x} + x^x - 2=0
\implies& a^{2 x} + x^x = 2\\
\end{align*}
We can see by inspection that the derived limit below is valid.
\begin{align*}
&\lim_{x \to 0}a^{2 x} + x^x\space  =1+1=2\\
\text{ therefore}&\\
&\lim_{x\to 0}\dfrac{a^{2x}\space -2}{x^x}=-1 
\end{align*}
Another approach is using simple substitution , permitted here because there is no possible division by zero.
$$\lim_{x\to 0}\dfrac{a^{2x}\space -2}{x^x}
=\frac{2^0-2}{0^0}=\frac{-1}{1}=-1$$
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