Calculate the limit of $(1+x2^x)/(1+x3^x)$ to the power $1/x^2$ when $x\to 0$ I have a problem with this: $\displaystyle \lim_{x \rightarrow 0}{\left(\frac{1+x2^x}{1+x3^x}\right)^\frac{1}{x^2}}$. 
I have tried to modify it like this: $\displaystyle\lim_{x\rightarrow 0}{e^{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}}$ and then calculate the limit of the exponent: $\displaystyle \lim_{x\rightarrow 0}{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}$. 
But I don't know what to do next. Any ideas?
 A: $\ln(1+x2^x)=\ln(1+x(1+x\ln2))+o(x^2)$
A: If you want to do it with l'Hôpital's theorem, it's better trying first with
\begin{align}
\lim_{x\to0}\frac{\log(1+xa^x)-x}{x^2}
&\overset{\mathrm{(H)}}{=}\lim_{x\to0}\frac{\dfrac{a^x+xa^x\log a}{1+xa^x}-1}{2x}\\[2ex]
&=\lim_{x\to0}\frac{a^x(x\log a-x)+(a^x-1)}{2x(1+xa^x)}\\[2ex]
&=\lim_{x\to0}\frac{a^x}{1+xa^x}\frac{\log a-1}{2}+
   \lim_{x\to0}\frac{1}{2(1+xa^x)}\frac{a^x-1}{x}\\[2ex]
&=\log a-\frac{1}{2}
\end{align}
Just note that $\lim_{x\to0}(a^x-1)/x=\log a$ is the derivative of $x\mapsto a^x$ at $0$.
Thus
$$
\lim_{x\to0}\frac{\log(1+xa^x)-\log(1+xb^x)}{x^2}=
\lim_{x\to0}\frac{\bigl(\log(1+xa^x)-x\bigr)-\bigl(\log(1+xb^x)-x\bigr)}{x^2}
$$
which is $\log a-\log b$.
Without l'Hôpital,
$$
\lim_{x\to0}\frac{\log(1+xa^x)-x}{x^2}=
\lim_{x\to0}\frac{xa^x-x^2a^{2x}/2+o((xa^x)^2)-x}{x^2}=
\lim_{x\to0}\left(\frac{a^x-1}{x}-\frac{a^{2x}}{2}\right)
$$
Motivation. Why trying with $-x$? Because of the symmetry in the given limit and because $\log(1+xa^x)\sim xa^x-x^2a^{2x}$, so subtracting $x$ gives a second order infinitesimal.
A: let $f(x)=\ln\left(\frac{1+x2^x}{1+x3^x}\right)$ and $g(x)=x^2$ then we have $0/0$ for $x$ tends to $0$ you must calculate $$\frac{f'(x)}{g'(x)}$$ and look if the limit exists.
you will get this here $$1/2\,{\frac {{2}^{x}\ln  \left( 2 \right) {3}^{x}{x}^{2}-{2}^{x}{3}^{x
}\ln  \left( 3 \right) {x}^{2}+x{2}^{x}\ln  \left( 2 \right) -x{3}^{x}
\ln  \left( 3 \right) +{2}^{x}-{3}^{x}}{ \left( 1+x{2}^{x} \right) 
 \left( 1+x{3}^{x} \right) x}}
$$
and now L'Hospital one more times
the next quotient is $${\frac {{2}^{x} \left( \ln  \left( 2 \right)  \right) ^{2}{3}^{x}{x}^{
2}+2\,{2}^{x}\ln  \left( 2 \right) {3}^{x}x-{2}^{x}{3}^{x} \left( \ln 
 \left( 3 \right)  \right) ^{2}{x}^{2}-2\,{2}^{x}{3}^{x}\ln  \left( 3
 \right) x+2\,{2}^{x}\ln  \left( 2 \right) +x{2}^{x} \left( \ln 
 \left( 2 \right)  \right) ^{2}-2\,{3}^{x}\ln  \left( 3 \right) -x{3}^
{x} \left( \ln  \left( 3 \right)  \right) ^{2}}{2\,{2}^{x}\ln  \left( 
2 \right) {3}^{x}{x}^{3}+2\,{2}^{x}{3}^{x}\ln  \left( 3 \right) {x}^{3
}+2\,{2}^{x}\ln  \left( 2 \right) {x}^{2}+6\,{2}^{x}{3}^{x}{x}^{2}+2\,
{3}^{x}\ln  \left( 3 \right) {x}^{2}+4\,x{2}^{x}+4\,x{3}^{x}+2}}
$$ and here you can set $x=0$ the result it $$\ln(2)-\ln(3)$$
A: This question seems good.
We have $$ \lim \limits_{x \to 0}{( \frac {1+x2^x}{1+x3^x})}^{1/{x^2}}$$
This is of the form $ \lim \limits_{x \to 0} {f(x)}^{g(x)}$ where $\lim \limits_{x \to 0}f(x) = 1$ and $\lim \limits_{x \to 0}g(x) = \infty$. We can modify the given limit as $ \lim\limits_{x \to 0} (1+ p(x) -1)^{g(x)}$ where $\lim \limits_{x \to 0}p(x) = 0$. This is nothing but equal to $e^{\lim \limits_{x \to 0} {(p(x)-1)g(x)} }$.
Hence we can write the given problem as $$=\lim \limits_{x \to 0} {(1+ \frac{x2^x - x3^x}{1+x3^x})}^{1/x^2}$$
$$=e^{\lim \limits_{x \to 0}\frac{x (2^x - 3^x)}{x^2 (1 + x3^x)}}$$. Now we use L hospital's rule and put $x=0$ in the limit $x$ tending to $0$ and we obtain
$$=e^{ln {\frac{2}{3}}}$$. $$= \frac{2}{3}$$
A: Since
$$
2^x=1+\log(2)x+O\!\left(x^2\right)\\
\Downarrow\\
\log\left(1+x2^x\right)=x+\left(\log(2)-\tfrac12\right)x^2+O\!\left(x^3\right)
$$
and
$$
3^x=1+\log(3)x+O\!\left(x^2\right)\\
\Downarrow\\
\log\left(1+x3^x\right)=x+\left(\log(3)-\tfrac12\right)x^2+O\!\left(x^3\right)
$$
Therefore,
$$
\frac1{x^2}\log\left(\frac{1+x2^x}{1+x3^x}\right)=\log\left(\frac23\right)+O\!\left(x\right)
$$
Applying $e^x$ gives
$$
\left(\frac{1+x2^x}{1+x3^x}\right)^{1/x^2}=\frac23\left(1+O(x)\right)
$$
A: As usual we take logs here because we have an expression of the form $\{f(x) \} ^{g(x)} $. We can proceed as follows
\begin{align}
\log L&=\log\left\{\lim_{x\to 0}\left(\frac{1+x2^{x}}{1+x3^{x}}\right)^{1/x^{2}}\right\}\notag\\
&=\lim_{x\to 0}\log\left(\frac{1+x2^{x}}{1+x3^{x}}\right)^{1/x^{2}}\text{ (via continuity of log)} \notag\\
&=\lim_{x\to 0}\frac{1}{x^{2}}\log\frac{1+x2^{x}}{1+x3^{x}}\notag\\
&=\lim_{x\to 0}\frac{1}{x^{2}}\cdot\dfrac{\log\dfrac{1+x2^{x}}{1+x3^{x}}}{\dfrac{1+x2^{x}}{1+x3^{x}}-1}\cdot\left(\frac{1+x2^{x}}{1+x3^{x}}-1\right)\notag\\
&=\lim_{x\to 0}\frac{1}{x^{2}}\cdot\frac{x(2^{x}-3^{x})}{1+x3^{x}}\notag\\
&=\lim_{x\to 0}\frac{2^{x}-3^{x}}{x}\notag\\
&=\lim_{x\to 0}\frac{2^{x}-1}{x}-\frac{3^{x}-1}{x}\notag\\
&=\log 2-\log 3\notag\\
&=\log(2/3)\notag
\end{align} 
and hence $L=2/3$. We have used the following standard limits here $$\lim_{x\to 1}\frac{\log x} {x-1}=1,\,\lim_{x\to 0}\frac{a^{x}-1}{x}=\log a$$ There is no need of more powerful tools like L'Hospital's Rule or Taylor's theorem. 
