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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...

Did
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Vinay5forPrime
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    Did you check the hypotheses of L'Hopital rule before applying it to $\lim\limits_{x\to0}\frac{x\log x}{\cos x}$? (You might not be responsible for having been led to believe this anecdotal tool is crucial and the first way to go to compute any limit but it is your responsability to check its hypotheses.) – Did Nov 29 '15 at 08:41
  • so i can stop at this step right? – Vinay5forPrime Nov 29 '15 at 08:42
  • so the answer is zero – Vinay5forPrime Nov 29 '15 at 08:43
  • The ingeniosity deployed to use L'H at all cost (by two answerers so far) is either admirable or disquieting (or both). – Did Nov 29 '15 at 08:51
  • can we put the limit at $\frac{x\log x}{\cos x}$ the answer will be zero straight away – Vinay5forPrime Nov 29 '15 at 08:54
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    Indeed, $$\tan x\cdot\log x=\frac{\sin x}{x}\cdot\frac1{\cos x}\cdot(x\log x)\to1\cdot1\cdot0.$$ – Did Nov 29 '15 at 08:57

4 Answers4

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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 $$L=-\lim_{x\to 0} \frac{\tfrac1x}{\tfrac{1}{x^2}\cos x+\tfrac1x\sin x}$$ $$=-\lim_{x\to 0} \frac{x}{\cos x+x\sin x}$$

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

rkm0959
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\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}

Hamit
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$$\lim_{x\to 0}\tan(x)\ln(x)=\lim_{x\to 0}\frac{\ln(x)}{\frac{1}{\tan(x)}}=\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\ln(x)}{\frac{\text{d}}{\text{d}x}\left(\frac{1}{\tan(x)}\right)}=$$ $$\lim_{x\to 0}\frac{\frac{1}{x}}{-\csc^2(x)}=\lim_{x\to 0}-\frac{\sin^2(x)}{x}=-\left(\lim_{x\to 0}\frac{\sin^2(x)}{x}\right)=$$ $$-\left(\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin^2(x)}{\frac{\text{d}}{\text{d}x}x}\right)=-\left(\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{1}\right)=-2\lim_{x\to 0}\cos(x)\sin(x)=$$ $$-2\cos(0)\sin(0)=-2\cdot 1\cdot 0=0$$

Jan Eerland
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