How can I determine $$ \lim_{x\to 0} \frac{\sqrt[2]{\cos(x)}+\sqrt[3]{\cos(x)}+\ldots+\sqrt[n+1]{\cos(x)}-n}{(e^x -1)^2} $$ without using L'Hôpital's rule?
Does it equal infinity?
I know about this remarkable limit that leaves the bottom part of the equation to an $x^2$, it being 0, makes it equal to infinite if the sum above is different than 0. Am I mistaking somewhere?
Here's a photo of the exercise to make it more clear:IMAGE
For limits without L'Hopital, the trick is always to multiply and divide by terms until you get known limits and make use of the property $\lim a_n b_n = \lim a_n \lim b_n$ when both exist. Here rewrite the limit as
$$\lim_{x\to 0}\left(\frac{x}{e^x-1}\right)^2\cdot\lim_{x\to 0}\frac{\sqrt[2]{\cos x}+\cdots \sqrt[n+1]{\cos x} - n}{x^2}$$
The limit on the left is a known limit, which can be proven to be $1$ by squeeze theorem. Next, split up the limits on the right
$$= \lim_{x\to 0}\frac{\sqrt[2]{\cos x}-1}{x^2}+\cdots+\lim_{x\to 0}\frac{\sqrt[n+1]{\cos x}-1}{x^2}$$
By geometric series, we know that
$$(r-1)(1+\cdots+r^{N-1}) = r^N-1$$
So for each $k$th root of $\cos x$, multiply and divide by the geometric series up to $k-1$ (Denote each sublimit by $L_k$)
$$L_k \equiv \lim_{x\to 0}\frac{\sqrt[k]{\cos x}-1}{x^2}\cdot\frac{1+\cdots+(\sqrt[k]{\cos x})^{k-1}}{1+\cdots+(\sqrt[k]{\cos x})^{k-1}} = \lim_{x\to 0}\frac{\cos x - 1}{x^2}\cdot\frac{1}{1+\cdots+(\sqrt[k]{\cos x})^{k-1}}$$
$$=\lim_{x\to 0}\left(\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2\cdot\lim_{x\to 0}\frac{-\frac{1}{2}}{1+\cdots+(\sqrt[k]{\cos x})^{k-1}}$$
The limit on the left is another known limit which equals $1$, and the limit on the right has no more singularities so we are free to take the limit by continuity
$$\implies L_k = -\frac{1}{2k}$$
which means the original limit was
$$-\frac{1}{2}\sum_{k=2}^{n+1}\frac{1}{k} = \frac{1-H_{n+1}}{2}$$
where $H_n$ is the $n$th Harmonic number.