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I kinda got stuck on this one...

How do I find $$\lim_{x\to 1}\frac{\tan(\ln(x))}{\ln(x)}$$

What's the technique for solving limits of 0/0?

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4 Answers 4

up vote 6 down vote accepted

Observe that $$\lim_{x\to 1}\frac{\tan (\ln x)}{\ln x}=\lim_{u\to 0}\frac{\tan u}{u}$$ via the substitution $u=\ln x$. Then, $$\lim_{u\to 0}\frac{\tan u}{u}=\lim_{u\to 0}\frac{\sin u}{u\cos u}=1$$ since $$\lim_{u\to 0}\frac{\sin u}{u}=1$$ The general technique for solving limits $0/0$ is with the De L'Hospital Rule

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Hint: See that when $x\to1$ then $\ln(x)\to 0$ and that $\lim_{t\to 0}\frac{\tan(t)}{t}=1$

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$+1^+\quad $ Short and sweet hint! $\quad\ddot\smile\quad$ –  amWhy Apr 14 '13 at 1:27

Besides the above approaches, we can use Taylor series:

$$\tan\log x=\log x+\frac{\log^3x}{3}+\frac{2\log^5x}{15}+...\,\,,\,\text{for}\,|\log x|<\frac{\pi}{2}\Longrightarrow$$

$$\frac{\tan\log x}{\log x}=1+\frac{\log^2x}{3}+\mathcal O(\log^4x)\xrightarrow[x\to 1]{} 1$$

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Nice, Nice approach. +1 –  Babak S. Dec 22 '12 at 17:23

Use L'Hôpital's rule: $$ \lim_{x\to 1}\frac{\tan(\ln(x))}{\ln(x)}=\lim_{x\to 1}\frac{\sec^2(\ln(x))/x}{1/x}= \lim_{x\to 1}{\sec^2(\ln(x))}={\frac1{\cos^2(\ln(1))}}=1 $$

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