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 $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$.
 A: We don't need to make any manipulation, by limit quotient theorem we have that for $a\neq 0$ and $x>0$
$$\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=\dfrac{\lim_{x\rightarrow0}\left(a^{2x}-2\right)}{\lim_{x\rightarrow0}x^x}=\frac{-1}1=-1$$
indeed

*

*$a^{2x}-2\to1-2=-1$

*$x^x=e^{x \log x} \to e^0=1$
for the latter refer also to

*

*A simple proof of $\lim_{n\to \infty} \frac{\ln n}{n}=0$ for students of a high school
A: I think it is enough to consider $x>0$ due to quantities like $x^x$. In this case $x\ln x \to 0$ as $x\to 0+$. The result follows due to $x^x = \exp (x\ln x)$.
A: 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 $\displaystyle\lim_{x\to 0}\dfrac{1}{e^{x\cdot \ln x}}=\dfrac{1}{e^{(\; \lim_{x\to 0}x\cdot \ln x)}}=\dfrac{1}{e^{0}}=1$
A: 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.
$$\lim_{x\to 0}\dfrac{a^{2x}\space -2}{x^x}
=\frac{2^0-2}{0^0}=\frac{-1}{1}=-1$$
