Limit of $\tan x\cdot\log x$ when $x\to0$, of the type $0\cdot\infty$ I am solving the question
$$\lim_{x\to 0}\tan x \log x$$
I did it till here
$$\lim_{x\to 0}\frac{\log x}{\cot x}=\lim_{x\to 0}\frac{\log x \sin x}{\cos x}$$
$$\lim_{x\to 0}\frac{\log x}{\cos x}\frac{\sin x}{x}*x$$
$$\lim_{x\to 0}\frac{x\log x}{\cos x}$$
$$\lim_{x\to 0}\frac{x\frac{1}{x}+\log x}{-\sin x}$$
$$\lim_{x\to 0}\frac{1+\log x}{-\sin x}$$
$$\lim_{x\to 0}\frac{\frac{1}{x}}{-\cos x}$$
I got stuck here some one please help...
 A: 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
$$L=-\lim_{x\to 0} \frac{\tfrac1x}{\tfrac{1}{x^2}\cos x+\tfrac1x\sin x}$$
$$=-\lim_{x\to 0} \frac{x}{\cos x+x\sin x}$$
A: 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 $$ \lim_{x\to 0} \frac{\frac{1}{x}}{\frac{-x\sin x-\cos x}{x^2}} = \lim_{x \to 0} \frac{x}{-\cos x - x \sin x}=0$$
A: \begin{eqnarray}
\lim_{x\to 0}\tan x \log x &=& \lim_{x\to 0}\dfrac{\log(x)}{\cot(x)}\\
&=& \lim_{x\to 0} \dfrac{\frac{1}{x}}{-\csc^{2}(x)}\\
&=& \lim_{x\to 0} -\dfrac{\sin^{2}(x)}{x}\\
&=& \lim_{x\to 0} -\sin(x)\times \dfrac{\sin(x)}{x}=0
\end{eqnarray}
A: $$\lim_{x\to 0}\tan(x)\ln(x)=\lim_{x\to 0}\frac{\ln(x)}{\frac{1}{\tan(x)}}=\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\ln(x)}{\frac{\text{d}}{\text{d}x}\left(\frac{1}{\tan(x)}\right)}=$$
$$\lim_{x\to 0}\frac{\frac{1}{x}}{-\csc^2(x)}=\lim_{x\to 0}-\frac{\sin^2(x)}{x}=-\left(\lim_{x\to 0}\frac{\sin^2(x)}{x}\right)=$$
$$-\left(\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin^2(x)}{\frac{\text{d}}{\text{d}x}x}\right)=-\left(\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{1}\right)=-2\lim_{x\to 0}\cos(x)\sin(x)=$$
$$-2\cos(0)\sin(0)=-2\cdot 1\cdot 0=0$$
