# Limits

• Aug 12th 2009, 05:34 PM
lehder
Limits
HI,

I didn't knox how to calculate the following limits:

1)-$\displaystyle \lim_{x\to \frac{\pi}{4}} f(x)=\frac{tanx-1}{\sqrt{2}cosx-1}$.

2)-$\displaystyle \lim_{x\to 0} f(x)=\frac{cosx+cos2x-2}{cos3x+sinx-1}$

And Thank's for anyway.
• Aug 12th 2009, 06:10 PM
pickslides
Quote:

Originally Posted by lehder
HI,

I didn't knox how to calculate the following limits:

1)-$\displaystyle \lim_{x\to \frac{\pi}{4}} f(x)=\frac{tanx-1}{\sqrt{2}cosx-1}$.

Have you herd of L'hospital's rule?

For $\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)}$

if $\displaystyle \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0$

then $\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}$
• Aug 12th 2009, 06:21 PM
lehder
Quote:

Originally Posted by pickslides
Have you herd of L'hospital's rule?

For $\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)}$

if $\displaystyle \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0$

then $\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}$

Thank you very much, i find that 1) = -2
• Aug 12th 2009, 06:25 PM
pickslides
I agree
• Aug 12th 2009, 06:37 PM
lehder
Quote:

Originally Posted by pickslides
I agree

And the second is egal = 0