How to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$

Question

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 $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$.

Answer 1

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 $x^x = \exp (x\ln x)$.

Answer 2

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

Answer 3

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.

$$\lim_{x\to 0}\dfrac{a^{2x}\space -2}{x^x} =\frac{2^0-2}{0^0}=\frac{-1}{1}=-1$$

Answer 4

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 $\displaystyle\lim_{x\to 0}\dfrac{1}{e^{x\cdot \ln x}}=\dfrac{1}{e^{(\; \lim_{x\to 0}x\cdot \ln x)}}=\dfrac{1}{e^{0}}=1$