Finding a limit without L'Hopital's rule or any series

Question

The limit is

$$\lim _{x\to 0 }\left(\frac{1 + x2^x}{1 + x3^x}\right)^\frac{1}{x^2}$$

I have no idea what to do. There are tons of exercises like this in my textbook and I was hoping if you could show me how to solve this one I would be able to solve others by myself.

I think I should use this $\lim _{x\to 0 }\left(\frac{a^x - 1}{x}\right) = \ln a$ but I don't know how. If you could give me some hints or solutions or even websites where there are solved problems like this that would be great help.

Thanks in advance!

Answer 1

Using $\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$, you can conclude that $$a^x=1+x\ln a+o(x).$$ Now substitute this result in the limit and try to express it in terms of something like $\lim_{x\to 0} (1+ax^2)^\frac{1}{x^2}$ (which is equal to $e^a$) and you are done.

Answer 2

\begin{align*} \left(\dfrac{1+x2^{x}}{1+x3^{x}}\right)^{\frac{1}{x^{2}}}&=\left(1+\dfrac{x3^{x}((\frac{2}{3})^{x}-1)}{1+x3^{x}}\right)^{\frac{1}{x^{2}}}\\ &=\left(1+\dfrac{x3^{x}((\frac{2}{3})^{x}-1)}{1+x3^{x}}\right)^{\dfrac{1+x3^{x}}{x3^{x}((\frac{2}{3})^{x}-1)}\cdot\left(\dfrac{x3^{x}((\frac{2}{3})^{x}-1)}{1+x3^{x}}\right)\cdot\dfrac{1}{x^{2}}}, \end{align*} while we see that \begin{align*} \left(\dfrac{x3^{x}((\frac{2}{3})^{x}-1)}{1+x3^{x}}\right)\cdot\dfrac{1}{x^{2}}=\dfrac{3^{x}}{1+x3^{x}}\cdot\dfrac{(\frac{2}{3})^{x}-1}{x}\rightarrow\log\left(\dfrac{2}{3}\right), \end{align*} so the limit is then $e^{\log(\frac{2}{3})}=\dfrac{2}{3}$.

Answer 3

$\begin{gathered} \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + x{2^x}}}{{1 + x{3^x}}}} \right)^{\frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + x{3^x} + x\left( {{2^x} - {3^x}} \right)}}{{1 + x{3^x}}}} \right)^{\frac{1}{{{x^2}}}}} \\ = \mathop {\lim }\limits_{x \to 0} {\left( {1 + \frac{{x\left( {{2^x} - {3^x}} \right)}}{{1 + x{3^x}}}} \right)^{\frac{{1 + x{3^x}}}{{x\left( {{2^x} - {3^x}} \right)}}.\frac{{\left( {{2^x} - {3^x}} \right)}}{{\left( {1 + x{3^x}} \right)x}}}} \hfill \\\\ = \mathop {\lim }\limits_{x \to 0} {e^{\frac{{\left( {{2^x} - {3^x}} \right)}}{{\left( {1 + x{3^x}} \right)x}}}} = \mathop {\lim }\limits_{x \to 0} {e^{\frac{{\left( {{2^x} - 1 - {3^x} + 1} \right)}}{{\left( {1 + x{3^x}} \right)x}}}} = {e^{\ln 2 - \ln 3}} = \frac{2}{3} \hfill \\ \end{gathered}$

It's easy with $\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = e$