# Thread: evaluate limit

1. ## evaluate limit

use L'Hopitals Rule
as the limit x approaches 0 from the right side
tanxlnx

i got as far as this lnx/(1/tanx) and tried to differentiate but coudnt get anything that sounded right

2. Hello, fgkl8!

Use L'Hopital's Rule: . $\lim_{x\to0^+} \tan x\ln x$

Since the limit goes to: . $0\cdot(\text{-}\infty)$, we can apply L'Hopital.

We have: . $\tan x \ln x \:=\:\frac{\ln x}{\cot x}$

Apply L'Hopital: . $\frac{\frac{1}{x}}{-\csc^2\!x} \;=\;-\frac{\sin^2\!x}{x} \;=\;-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)$

Therefore: . $\lim_{x\to0^+}\bigg[-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)\bigg] \;=\;-(1)(\infty) \;=\;-\infty$

3. Originally Posted by Soroban
Hello, fgkl8!

Since the limit goes to: . $0\cdot(\text{-}\infty)$, we can apply L'Hopital.

We have: . $\tan x \ln x \:=\:\frac{\ln x}{\cot x}$

Apply L'Hopital: . $\frac{\frac{1}{x}}{-\csc^2\!x} \;=\;-\frac{\sin^2\!x}{x} \;=\;-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)$

correction ...

Therefore: . $\lim_{x\to0^+}\textcolor{red}{\bigg[-\left(\frac{\sin x}{x}\right)\left(\frac{\sin{x}}{1}\right)\bigg] \;=\;-(1)(0) = 0 }$

...

4. Originally Posted by Soroban
Hello, fgkl8!

Since the limit goes to: . $0\cdot(\text{-}\infty)$, we can apply L'Hopital.

We have: . $\tan x \ln x \:=\:\frac{\ln x}{\cot x}$

Apply L'Hopital: . $\frac{\frac{1}{x}}{-\csc^2\!x} \;=\;-\frac{\sin^2\!x}{x} \;=\;-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)$
EDIT: Was just a typo in. missed that a sin was missing just noticed the (sin/x)(1/x) bit.

Therefore: . $\lim_{x\to0^+}\bigg[-\left(\frac{\sin x}{x}\right)\left(\frac{1}{x}\right)\bigg] \;=\;-(1)(\infty) \;=\;-\infty$

i dont agree with the last line, lim(ab)=lim (a)lim(b) if both limits exist.
applying l'hopitals once again gives

lim 2sinxcosx/1 as x goes to 0

which is 0

5. for $x\ne0$ we have $\tan (x)\ln (x)=\frac{\tan x}{x}\cdot x\ln (x),$ and we can compute its limit.