Limit of $\tan x\cdot\log x$ when $x\to0$, of the type $0\cdot\infty$

Solution 1:

You cannot apply l'Hospital rule to the limit $$\lim_{x\to 0} \frac{x\log x}{\cos x}$$

because it's not in the form $\frac00$ or $\frac{\infty}{\infty}$. Instead try to rewrite the limit as $$L=\lim_{x\to 0} \frac{\log x}{\tfrac1x\cos x}$$ and apply the l'Hospital rule as the following $$L=-\lim_{x\to 0} \frac{\tfrac1x}{\tfrac{1}{x^2}\cos x+\tfrac1x\sin x}$$ $$=-\lim_{x\to 0} \frac{x}{\cos x+x\sin x}$$

Solution 2:

I will start from $$\lim_{x \to 0} \frac{x\ln x}{\cos x}$$ Change this into $$\lim_{x \to 0} \frac{\ln x}{\frac{\cos x}{x}}$$

Use L'Hopital (since both the numerator and denominator $\rightarrow \infty$)

We get $$ \lim_{x\to 0} \frac{\frac{1}{x}}{\frac{-x\sin x-\cos x}{x^2}} = \lim_{x \to 0} \frac{x}{-\cos x - x \sin x}=0$$