$\lim_{x \to 0} (e^x - 1)/x^8$ . I want to know how to solve this limit problem with Maclaurin's series expansion of only $e^x$ in the neighborhood of $x=0$.
 A: You can certainly apply l'Hôpital here, provided you distinguish between the limits from the right and from the left. Since
$$
\lim_{x\to0^+}\frac{e^x}{8x^7}=\infty
$$
you can certainly conclude that
$$
\lim_{x\to0^+}\frac{e^x-1}{x^8}=\infty
$$
Similarly
$$
\lim_{x\to0^-}\frac{e^x}{8x^7}=-\infty
$$
and therefore
$$
\lim_{x\to0^-}\frac{e^x-1}{x^8}=-\infty
$$
Hence the two-sided limit doesn't exist.
What about Taylor (or Maclaurin, if you prefer to call it this way)? The Taylor expansion is $e^x-1=x+o(x)$ so you have, from the right,
$$
\lim_{x\to0^+}\frac{e^x-1}{x^8}=\lim_{x\to0}\frac{x+o(x)}{x^8}=\infty
$$
(you get $\infty$ because you know that the function is positive in a right punctured neighborhood of $0$). Similarly for the limit from the left and you get the same result as before.
A: The limit does not exist. $(e^x-1)/x^8$ approaches infinity as $x$ approaches $0$ from the positive end, but $(e^x-1)/x^8$ approaches negative infinity as $x$ approaches $0$ from the negative end. L'Hôpital's rule states that if $f(x) \to 0$ and $g(x) \to 0$ as $x$ approaches $a$, then
$$
\lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f'(x)}{g'(x)}
$$
provided that
$$
\lim_{x \to a}\frac{f'(x)}{g'(x)}
$$
exists. In this case, that limit does not exist, and so L'Hôpital is not applicable.
A: $$\frac{e^x-1}{x^8}=\frac{1+x+o(x^2)-1}{x^8}.$$
You can conclude.
