# Thread: Need a method for solving this..

1. ## [SOLVED]Need a method for solving this..

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

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

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

What am I doing wrong?

2. You did not take the derivative of the denominator while using Lhopital's rule!

3. Aren't you also supposed to take the derivative of the denominator?

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

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

You just replaced $\displaystyle 1-cos^2(x)$ with $\displaystyle sin^2(x)$
Right sorry..

I meant

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

Even then I get stuck, can someone point me in the right direction?

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

use Lhopitals rule to get:

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

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

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

use Lhopitals rule once again to get the answer.

6. This goes to $\displaystyle \displaystyle \frac{0}{0}$, so you can use L'Hospital's Rule again. Also note that $\displaystyle \displaystyle 2\sin{x}\cos{x} = \sin{2x}$, which will make differentiating the denominator easier...

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

use Lhopitals rule to get:

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

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

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

use Lhopitals rule once again to get the answer.
$\displaystyle \displaymode = \dfrac{(x-1)(-\sin(x))+ \cos(x)(1)}{(x-1)cos(2x)(2)+sin(2x)(1)}$

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

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

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

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

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

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

Otherwise there are no serious errors. good job!

9. Originally Posted by harish21
The denominator should be $\displaystyle 2x\cos(2x)-2\cos(2x)+\sin(2x)$. 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.

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

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

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

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

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

I understand the sin(2x) = 2sin(x)cos(x) part.

10. $\displaystyle \dfrac{1}{x-1}+\cos(x)$

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

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

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

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

$\displaystyle = \dfrac{1+\cos(x)(x-1)}{x-1}$
You are amazing. Thanks

12. Actually, in this case it's easier to use L'Hospital's Rule the second time BEFORE doing any algebraic manipulation...

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

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

The limit is now evaluated through direct substitution.