Need a method for solving this..

• Feb 10th 2011, 09:14 PM
qleeq
[SOLVED]Need a method for solving this..
$\lim_{x \to 0} \frac{ln(1-x) + sin(x)}{1-cos^2(x)}$

Tried using L'Hopital and I know that's what you have to use but I still get stuck here..

$\lim_{x \to 0} \frac{\frac{1}{x-1}+cos(x)}{sin^2(x)}$

What am I doing wrong?
• Feb 10th 2011, 09:17 PM
harish21
You did not take the derivative of the denominator while using Lhopital's rule!
• Feb 10th 2011, 09:20 PM
guildmage
Aren't you also supposed to take the derivative of the denominator?

You just replaced $1-cos^2(x)$ with $sin^2(x)$
• Feb 10th 2011, 09:24 PM
qleeq
Quote:

Originally Posted by guildmage
Aren't you also supposed to take the derivative of the denominator?

You just replaced $1-cos^2(x)$ with $sin^2(x)$

Right sorry..

I meant

$\lim_{x \to 0} \frac{\frac{1}{x-1}+cos(x)}{-2cos(x)sin(x)}$

Even then I get stuck, can someone point me in the right direction?
• Feb 10th 2011, 09:34 PM
harish21
$\displaystyle \lim_{x \to 0} \dfrac{\ln(1-x)+sin(x)}{\sin^2(x)}$.

use Lhopitals rule to get:

$\displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{x-1}+\cos(x)}{2\sin(x)\cos(x)}$

$\displaystyle = \lim_{x \to 0} \dfrac{\dfrac{1+\cos(x)(x-1)}{x-1}}{\sin(2x)}$

$\displaystyle = \lim_{x \to 0}\dfrac{1+(x-1)\cos(x)}{(x-1)\sin(2x)}$

use Lhopitals rule once again to get the answer.
• Feb 10th 2011, 09:35 PM
Prove It
This goes to $\displaystyle \frac{0}{0}$, so you can use L'Hospital's Rule again. Also note that $\displaystyle 2\sin{x}\cos{x} = \sin{2x}$, which will make differentiating the denominator easier...
• Feb 10th 2011, 10:15 PM
qleeq
Quote:

Originally Posted by harish21
$\displaystyle \lim_{x \to 0} \dfrac{\ln(1-x)+sin(x)}{\sin^2(x)}$.

use Lhopitals rule to get:

$\displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{x-1}+\cos(x)}{2\sin(x)\cos(x)}$

$\displaystyle \lim_{x \to 0} \dfrac{\dfrac{1+\cos(x)(x-1)}{x-1}}{\sin(2x)}$

$\displaystyle \lim_{x \to 0}\dfrac{1+(x-1)\cos(x)}{(x-1)\sin(2x)}$

use Lhopitals rule once again to get the answer.

$\displaymode = \dfrac{(x-1)(-\sin(x))+ \cos(x)(1)}{(x-1)cos(2x)(2)+sin(2x)(1)}$

$\displaymode = \dfrac{-\sin(x)+\sin(x)+\cos(x)}{2x\cos(x)-2\cos(x)+\sin(2x)}$

Then plugging it all in..
$-x\sin(0)+\sin(0)+\cos(0) = 0 + 0 + 1 = 1$

$2(0)\cos(0)-2\cos(0)+\sin(2(0)) = 0 - 2(1) + 0 = -2$

$\displaymode = -\frac{1}{2}$

• Feb 10th 2011, 10:16 PM
harish21
Quote:

Originally Posted by qleeq
$\displaymode = \dfrac{-\sin(x)+\sin(x)+\cos(x)}{2x\cos(x)-2\cos(x)+\sin(2x)}$

The denominator should be $2x\cos(2x)-2\cos(2x)+\sin(2x)$. You missed typing the 2s I guess.

Otherwise there are no serious errors. good job!
• Feb 10th 2011, 10:33 PM
qleeq
Quote:

Originally Posted by harish21
The denominator should be $2x\cos(2x)-2\cos(2x)+\sin(2x)$. You missed typing the 2s I guess.

Otherwise there are no serious errors. good job!

Must have accidentally removed them while I was editing.. Anyway, thanks a lot for your help. I had one last question though.

$\displaystyle \lim_{x \to 0} \dfrac{\dfrac{1+\cos(x)(x-1)}{x-1}}{\sin(2x)}$

$\displaystyle \lim_{x \to 0}\dfrac{1+(x-1)\cos(x)}{(x-1)\sin(2x)}$

Can you explain to me how he arrived at the firstequation from $\displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{x-1}+\cos(x)}{2\sin(x)\cos(x)}$?

I understand the second part, but the first part has me confused.

How did he get $\dfrac{\dfrac{1+cos(x)(x-1)}{(x-1)}}{sin(2x)}$ from $\dfrac{\dfrac{1}{x-1}+cos(x)}{2sin(x)cos(x)}$ ?

I understand the sin(2x) = 2sin(x)cos(x) part.
• Feb 10th 2011, 10:41 PM
harish21
$\dfrac{1}{x-1}+\cos(x)$

$= \dfrac{1}{x-1}+ \bigg(\dfrac{\cos(x)}{1} \times \dfrac{x-1}{x-1}\bigg)$

$= \dfrac{1+\cos(x)(x-1)}{x-1}$
• Feb 10th 2011, 10:42 PM
qleeq
Quote:

Originally Posted by harish21
$\dfrac{1}{x-1}+\cos(x)$

$= \dfrac{1}{x-1}+ \bigg(\cos(x) \times \dfrac{x-1}{x-1}\bigg)$

$= \dfrac{1+\cos(x)(x-1)}{x-1}$

You are amazing. Thanks :D
• Feb 10th 2011, 11:27 PM
Prove It
Actually, in this case it's easier to use L'Hospital's Rule the second time BEFORE doing any algebraic manipulation...

$\displaystyle \lim_{x \to 0}\frac{\frac{1}{x-1} + \cos{x}}{-2\sin{x}\cos{x}} = \lim_{x \to 0}\frac{(x-1)^{-1} + \cos{x}}{-\sin{2x}}$

$\displaystyle = \lim_{x \to 0}\frac{-(x-1)^{-2} - \sin{x}}{-2\cos{2x}}$ by L'Hospital's Rule.

The limit is now evaluated through direct substitution.