1. ## limit 2

lim 1-cos2x
x-->0 -------------=?
x^2+x.sinx

I applied L'Hospital but couldn't get rid of x.
I tried another way

1-cos2x=1-(cos^x-sin^x)=sin^2x

sin^2x
----------
x^2+xsinx

I tried to divide both sides by 2x to apply sandwich rule but I couldn't get anything.

2. ## Re: limit 2

Originally Posted by kastamonu
lim 1-cos2x
x-->0 -------------=?
x^2+x.sinx

I applied L'Hospital but couldn't get rid of x.
I applied L'Hospital's rule twice, and that was enough. It works out that way if you persevere.

3. ## Re: limit 2

2cosx/2x+sinx+xcosx I got.

And found the answer as 1/2 but book says 1. I made a mistake somewhere.

4. ## Re: limit 2

Originally Posted by kastamonu
2cosx/2x+sinx+xcosx <----- You need grouping symbols around the denominator (whether or not this expression is correct)
to show what you intend.

I got.

And found the answer as 1/2 but book says 1. I made a mistake somewhere.
Realize that there is supposed to be limit notation in front of each of the following lines that have expressions:

$\displaystyle \dfrac{1 - cos(2x)}{x^2 + xsin(x)}$

L'Hospital's rule (with the limit understood being applied):

$\displaystyle \dfrac{sin(2x)*2}{2x + sin(x) + xcos(x)}$

$\displaystyle \dfrac{2sin(2x)}{2x + sin(x) + xcos(x)}$

L'Hospital's rule (with the limit understood being applied) a second time:

$\displaystyle \dfrac{2cos(2x)*2}{2 + cos(x) + cos(x) - xsin(x)}$

$\displaystyle \dfrac{4cos(2x)}{2 + 2cos(x) - xsin(x)}$

Now, take the limit as x approaches 0 of the last expression.

5. ## Re: limit 2

Many Thanks.

May I also use Sandwich rule after applying L'Hospital? I am asking generally.