In the L'Hôpitals-section in my textbook I’m asked to solve:
$$\lim_{x\to 0^{+}}x^p(e^{-px}-\frac{1}{\ln x}-\frac{1}{x+1}) \; \text{when} \;p<0 $$
The only way forward I can imagine is to rewrite as
$$ lim_{x\to 0^{+}}\frac{e^{-px}-\frac{1}{\ln x}-\frac{1}{x+1}}{x^{-p}} $$
and solve with L'Hôpital's rule. Problem is that the numerator becomes increasingly complex if I do, so I don’t think the approach works. Are there any way I can rewrite the expression to avoid such complexities?
On
Let $p=-q, q>0$ and then we can write the given expression as $$\frac {((1+x)e^{qx}-1)\log x-1-x} {x^q(1+x)\log x} \tag{1}$$ Next note that $$\frac{(1+x)e^{qx}-1}{x}\to (q+1)$$ as $x\to 0^{+}$ and hence the numerator of $(1)$ tends to $-1$ and denominator tends to $0^{-}$. It follows that the desired limit is $\infty$.
The expression in $(\, )$ tends to $2$ as $x \to 0$ so the limit is $\infty$. L'Hopitals Rule is not applicable here.