When working through some exercises regarding limits, I stumbled across this one that puzzles me.
$$\lim_{x \to 0}\left(1 + x^3\right)^{\frac{1}{x-\sin(x)}} = e^6$$
Can anyone show me how you could work this out? I found the answer using wolframalpha. I know that x equals sin(x) as x tends to zero, so you raise a little bit more than one to an infinitely large power. I also know that
$$\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x = e.$$
But why does e pop up in the first equation? When x is raised to any other power than three the limit as x tends to zero is just one. What is it that makes three the special case?
Hint: As both "Student" and "Claude" have tried to say in the comments, a first step is to change the problem to one you are familiar with (Clause suggests using logarithms)
i.e. we have that $$\lim_{x\to0}(1+x^3)^{\frac{1}{x-\sin(x)}}=\lim_{x\to0}e^{\log(1+x^3)\cdot\frac{1}{x-\sin(x)}}=e^{\displaystyle\left(\lim_{x\to0}\frac{\log(1+x^3)}{x-\sin(x)}\right)}$$
Now we only have to calculate the limit $\lim_{x\to0}\frac{\log(1+x^3)}{x-\sin(x)}$ with one of the methods you have learned (l'Hopital/Taylor series/...).
The limit you mentioned $\left(\lim_{x\to\infty}(1+\frac1x)^x=e\right)$ seems to be a red herring.
This is because $\log(1+x^k)$ has a zero of order k in $x=0$ (And similarly $x-\sin x$ has a zero of order 3) (See also this question about order of zeroes), and thus for $k>3$ we have that $\log(1+x^k)$ has a zero of higher degree than $x-\sin x$. However, for $k<3$ we would have a problem, as the limit diverges.