1
$\begingroup$

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?

4 Answers 4

6

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

4

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

  • 0
    $+1^+\quad $ Short and sweet hint! $\quad\ddot\smile\quad$2013-04-14
3

Besides the above approaches, we can use Taylor series:

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

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

  • 0
    Nice, Nice approach. +12012-12-22
2

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 $