Limit problem.

• December 27th 2008, 08:54 PM
arvindshetty86
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.
• December 27th 2008, 08:59 PM
Chris L T521
Quote:

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?
• December 27th 2008, 08:59 PM
kalagota
Quote:

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
• December 27th 2008, 09:12 PM
arvindshetty86
yes Chris that's the question.Thanks for the help.
• December 27th 2008, 10:50 PM
Isomorphism
L'Hospitals
Quote:

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 (Thinking)

All this leads me to believe there is a typo (Nod)