Incorrect application of product rule for limits? I'm asked to compute $$\lim _{x \rightarrow 0^{+}} \frac{x^{\sin x}-1}{x}.$$
My textbook gives the following solution:
$$
\lim _{x \rightarrow 0^{+}} \frac{x^{\sin x}-1}{x}=\lim _{x \rightarrow 0^{+}} \frac{\mathrm{e}^{\sin x \log x}-1}{x} \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0^{+}} \underbrace{\mathrm{e}^{\sin x \log x}}_{\rightarrow 1}\left(\underbrace{\cos x \log x}_{\rightarrow-\infty}+\underbrace{\frac{\sin x}{x}}_{\rightarrow 1}\right)=-\infty .
$$
whereby H denotes the application of l'Hôpital's rule. I was wondering whether the last step is really  valid/rigorous because it uses the product (and sum-) rule for limits which, however, as stated in the textbook is only proven if the involved functions converge to some particular $L \in \mathbb{R}$.
 A: Yes, the last step is fully valid.  Suppose $\lim (f(x)+g(x))=C$ and $\lim f(x)=A$.  Then $\lim g(x)$ exists and $\lim g(x) = A-C$.  Conversely, if $\lim g(x)$ does not exist or is not finite but you know $\lim f(x) = A$, then $\lim(f(x)+g(x))$ cannot exist and be finite.
To show that the limit is in fact infinite, just note that for $x$ sufficiently close to $0^+, f(x)$ is bounded and since $g(x) \to \ - \infty$ the sum $f(x)+g(x)$ must get and stay arbitrarily large in magnitude.
Similar reasoning works for a product of functions.
A: Let's make the last step properly rigorous then. Let's prove the following proposition:

Suppose $f, g : (0, \lambda) \to \Bbb{R}$ such that $\lim_{x \to 0^+} f(x) = -\infty$ and $\lim_{x \to 0^+} g(x) = L \in \Bbb{R}$. Then $f(x) + g(x) \to -\infty$ as $x \to -\infty$.

Fix $M \in \Bbb{R}$. As $f(x) \to -\infty$, there exists some $\delta_1 > 0$ such that
$$0 < x < \delta_1 \implies f(x) < M - L - 1.$$
As $g(x) \to L$, there exists some $\delta_2 > 0$ such that
$$0 < x < \delta_2 \implies |g(x) - L| < 1 \implies g(x) < L + 1.$$
Let $\delta = \min\{\delta_1, \delta_2\} > 0$. Then
\begin{align*}
0 < x < \delta &\implies \cases{0 < x < \delta_1 \\ 0 < x < \delta_2} \\
&\implies \cases{f(x) < M - L - 1 \\ g(x) < L + 1} \\
&\implies f(x) + g(x) < M.
\end{align*}
This verifies that $\lim_{x \to 0^+} f(x) + g(x) = -\infty$ by definition.
