# Find the limit of a function

• Jun 26th 2009, 06:50 PM
razemsoft21
Find the limit of a function
can you help me solving this question
without using L'HOPITAL'S RULE

Attachment 11989
• Jun 26th 2009, 07:16 PM
Krizalid
just put $\displaystyle t=x-\frac\pi4.$
• Jun 26th 2009, 09:37 PM
o_O
If you know anything about derivatives ...

In general: $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$

Imagine: $\displaystyle a = \tfrac{\pi}{4}$ and $\displaystyle f(x) = \cos^2 (x)$
• Jun 30th 2009, 04:44 AM
razemsoft21
Quote:

Originally Posted by Krizalid
just put $\displaystyle t=x-\frac\pi4.$

OK , I put $\displaystyle t=x-\frac\pi4$
I got the following restul !!

http://www.mathhelpforum.com/math-he...1&d=1246365734
• Jun 30th 2009, 04:52 AM
malaygoel
Quote:

Originally Posted by razemsoft21
OK , I put $\displaystyle t=x-\frac\pi4$
I got the following restul !!

http://www.mathhelpforum.com/math-he...1&d=1246365734

use the identity$\displaystyle sin2x=2cos^2x-1$

and then the limit $\displaystyle \frac{sinx}{x}$
• Jun 30th 2009, 02:24 PM
mr fantastic
Quote:

Originally Posted by o_O
If you know anything about derivatives ...

In general: $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$

Imagine: $\displaystyle a = \tfrac{\pi}{4}$ and $\displaystyle f(x) = \cos^2 (x)$

Since the question is posted in PRE-Calculus it might be IF rather than if ....
• Jun 30th 2009, 04:16 PM
HallsofIvy
Quote:

Originally Posted by razemsoft21
OK , I put $\displaystyle t=x-\frac\pi4$
I got the following restul !!

http://www.mathhelpforum.com/math-he...1&d=1246365734

Use $\displaystyle cos(t+ \pi/4)= cos(t)cos(\pi/4)- sin(t)sin(\pi/4)$$\displaystyle = (\sqrt{2}/2)cos(t)- (\sqrt{2}/2)sin(t) so \displaystyle cos^2(t+ \pi/4)= \frac{1}{2}cos^2(t)- sin(t)cos(t)+ \frac{1}{2}sin^2(t) \displaystyle \frac{cos^2(t+ \pi/4)- \frac{1}{2}}{t}= \frac{1}{2}\frac{cos^2(t)-1}{t}- \frac{sin(t)}{t}cos(t)+ \frac{1}{2}\frac{sin(t)}{t}sin(t)$$\displaystyle = -\frac{1}{2}\frac{sin(t)}{t}sin(t)- \frac{sin(t)}{t}cos(t)+\frac{1}{2}\frac{sin(t)}{t} sin(t)$$\displaystyle = -\frac{sin(t)}{t}cos(t)$
• Jul 1st 2009, 12:35 PM
razemsoft21
Thanks
Quote:

Originally Posted by o_O
If you know anything about derivatives ...

In general: $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$

Imagine: $\displaystyle a = \tfrac{\pi}{4}$ and $\displaystyle f(x) = \cos^2 (x)$

Thank you for help, it works: (Clapping) razemsoft21 (Handshake) o_O

definition of f'(pi/4) = -2cos(pi/4).sin(pi/4) = -1
the same result when I use L'HOPITAL'S RULE