Calculus: a very serious L'Hopital's Rule problem Compute using L'Hopital's rule:
$$\lim_{x\to 0^+} \frac{\ln(x)}{1/\sin(x)}$$
I kept differentiating, but it's getting too long. How can I tackle this kind of problem?
Also, when I encounter a limit of the form $\infty \cdot 0$ and I want to make it eligible for L'Hopital's rule, should I transform it into $$\frac{\infty}{1/0}=\frac{\infty}{\infty}$$ or 
$$\frac{0}{1/\infty}=\frac{0}{0}$$
 A: Or without l'Hopital:
$$\frac{\ln x}{\frac1{\sin x}} = (x\ln x)\cdot \frac{\sin x}x,$$
where each factor has a well-known limit as $x\to0^+$.
A: You chose the poorer way to turn the $0\cdot\infty$ form into one to which l’Hospital’s rule applies.
$$\begin{align*}
\lim_{x\to 0^+}\ln x\sin x&=\lim_{x\to 0^+}\frac{\sin x}{1/\ln x}\\\\
&=\lim_{x\to 0^+}\frac{\cos x}{-(\ln x)^{-2}\cdot\frac1x}\\\\
&=-\lim_{x\to 0^+}\frac{(\ln x)^2\cos x}{1/x}\;.
\end{align*}$$
At this point you can try try applying l’Hospital’s rule again, but it’s clear that the numerator is going to be fairly messy. A better idea is to notice that $\lim\limits_{x\to 0^+}\cos x= 1$, so that
$$-\lim_{x\to 0^+}\frac{(\ln x)^2\cos x}{1/x}=-\left(\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}\right)\lim_{x\to 0^+}\cos x=-\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}\;;$$ this gets rid of the trig function. Now
$$\begin{align*}
-\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}&=-\lim_{x\to 0^+}\frac{\frac2x\ln x}{-1/x^2}\\\\
&=2\lim_{x\to 0^+}\frac{\ln x}{1/x}\\\\
&=2\lim_{x\to 0^+}\frac{1/x}{-1/x^2}\\\\
&=-2\lim_{x\to 0^+}x\\\\
&=0\;.
\end{align*}$$
It’s always a good idea to keep your eyes open for factors with known finite, non-zero limits, like the $\cos x$ above: generally speaking, it’s a good idea to simplify as much as possible the expression whose limit you’re taking.
A: When you use LHospital's rule
$$\lim_{x \to 0} \dfrac{\ln(x)}{\csc(x)} = \lim_{x \to 0} \dfrac{\left(\dfrac{d\ln(x)}{dx} \right)}{\left(\dfrac{d\csc(x)}{dx} \right)} = \lim_{x \to 0} \dfrac{\left(1/x \right)}{\left(-\cot(x) \csc(x) \right)} = -\lim_{x \to 0} \dfrac{\sin(x) \tan(x)}{x}\\ = - \left(\lim_{x \to 0} \dfrac{\sin(x)}{x} \right) \times \left(\lim_{x \to 0} \tan(x) \right)$$
