1. ## Limit problem.

Can any1 help me to find the solution for this problem

lim a sin square x b log cosx /4
x->0

equals to half i.e 1/2
Find a and b.

2. Originally Posted by arvindshetty86
Can any1 help me to find the solution for this problem

2
lim a sin^ x b log cosx 1
x->0 ------------------------- = ---
4 2

Find a and b.
$\lim_{x\to0}\frac{a\sin^2\left(x\right)\cdot b\log\left(\cos\left(x\right)\right)}{4}=\frac{1}{ 2}$

The way you posted the question is unclear and hard to interpret. Is this what you mean?

3. Originally Posted by arvindshetty86
Can any1 help me to find the solution for this problem

2
lim a sin^ x b log cosx 1
x->0 ------------------------- = ---
4 2

Find a and b.
hi can you fix your post? try using parenthesis and stuffs.. tnx

4. yes Chris that's the question.Thanks for the help.

5. ## L'Hospitals

Originally Posted by For Arvind by Chris
$\lim_{x\to0}\frac{a\sin^2\left(x\right)\cdot b\log\left(\cos\left(x\right)\right)}{4}=\frac{1}{ 2}$
Hello Arvind,

Can you recheck the question? Something is not right here...

Why would they keep constants like that (a and b) when they can group it and call it one constant ( like c = ab) ?

And secondly, a simple L'Hospital will show that:

$\lim_{x\to0}\sin^2 x\cdot \log\left(\cos x\right) = 0$

So I dont think a constant ab exists

All this leads me to believe there is a typo