Find the limit of a function?? Determine the following limit:
$$\lim\limits_{x\to0}\frac{(1+x\cdot2^x)^{\frac{1}{x^2}}}{(1+x\cdot3^x)^{\frac{1}{x^2}}}$$
 A: $$\lim_{x\to0}\frac{(1+x\cdot2^x)^{\frac{1}{x^2}}}{(1+x\cdot3^x)^{\frac{1}{x^2}}}$$
$$=\lim_{x\to0}\left(\dfrac{1+x\cdot2^x}{1+x\cdot3^x}\right)^{\dfrac1{x^2}} =\lim_{x\to0}\left(1+\dfrac{x(2^x-3^x)}{1+x\cdot3^x}\right)^{\dfrac1{x^2}}$$
$$=\left(\lim_{x\to0}\left(1+\dfrac{x(2^x-3^x)}{1+x\cdot3^x}\right)^{\dfrac{1+x\cdot3^x}{x(2^x-3^x)}}\right)^{\lim_{x\to0}\dfrac{\dfrac{x(2^x-3^x)}{1+x\cdot3^x}}{x^2}}$$
Setting $\dfrac{1+x\cdot3^x}{x(2^x-3^x)}=n$ which $\to\infty$
the inner limit converges to $e$ as $\displaystyle \lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$
For $\displaystyle\lim_{x\to0}\dfrac{\dfrac{x(2^x-3^x)}{1+x\cdot3^x}}{x^2}$
$=\displaystyle\frac1{\lim_{x\to0}(1+x\cdot3^x)}\cdot\left(\lim_{x\to0}\frac{2^x-1}x-\lim_{x\to0}\frac{3^x-1}x\right)$
$\displaystyle=1\cdot(\ln2-\ln 3)=\ln\frac23$
Now we know, $\displaystyle e^{\ln a}=a$ for real $a>0$
A: Let $y=\frac{(1+x\cdot2^x)^{\frac{1}{x^2}}}{(1+x\cdot3^x)^{\frac{1}{x^2}}}$
Then ln($y$) = $\dfrac{1}{x^2}(ln(1+x.2^x)-ln(1+x.3^x))$. Now to find the limit of the RHS as $x \rightarrow 0$ just apply L Hospital's rule.
A: You can also start just as suggested by voldemort and use the Taylor expansion of $\log (1+x a^x)$ which is $$\log (1+x a^x) \simeq x+x^2 \left(\log (a)-\frac{1}{2}\right)+x^3 \left(\frac{\log ^2(a)}{2}-\log
   (a)+\frac{1}{3}\right)+O\left(x^4\right)$$ Apply it for numerator and denominator and you will arrive to the result given by lab bhattacharjee.
