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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!

Britanica
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3 Answers3

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

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

HOANXA
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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.

Mostafa Ayaz
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  • Can I ask for a favor ? – Britanica May 01 '21 at 16:25
  • Sure. Do you wish for a full solution on OP?${}{}{}{}$ – Mostafa Ayaz May 01 '21 at 16:26
  • I got run over a by a car (I am not joking, I swear to God) and that's why I'm behind my mathematical analysis class. Could you recommend some resources where I can find what $o(x)$ is? – Britanica May 01 '21 at 16:30
  • It is okay :) The little-o notation roughly provides information about the order of changes of a function $f(x)$ with respect to $g(x)$. A simple definition is as follows$$f(x)=o(g(x))\iff \lim_{x\to 0}\frac{f(x)}{g(x)}=0.$$ Also you can find more details in https://mathworld.wolfram.com/Little-ONotation.html. – Mostafa Ayaz May 01 '21 at 16:35
  • Thank you so much. You're too kind :) – Britanica May 01 '21 at 16:36
  • Happy to help! And a final note: what I meant through using little-o notation, was to say that the term $a^x-1-x\ln a$ decays faster than linear by $x$. – Mostafa Ayaz May 01 '21 at 16:38
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\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 $e^{\log(\frac{2}{3})}=\dfrac{2}{3}$.

user284331
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