Find the limit of a function

**razemsoft21** · June 26th 2009, 06:50 PM

can you help me solving this question
without using L'HOPITAL'S RULE

**Krizalid** · June 26th 2009, 07:16 PM

just put $t=x-\frac\pi4.$

**o_O** · June 26th 2009, 09:37 PM

If you know anything about derivatives ...

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

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

**razemsoft21** · June 30th 2009, 04:44 AM

Originally Posted by Krizalid

just put $t=x-\frac\pi4.$

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

**malaygoel** · June 30th 2009, 04:52 AM

Originally Posted by razemsoft21

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

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

and then the limit $\frac{sinx}{x}$

**mr fantastic** · June 30th 2009, 02:24 PM

Originally Posted by o_O

If you know anything about derivatives ...

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

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

Since the question is posted in PRE-Calculus it might be IF rather than if ....

**HallsofIvy** · June 30th 2009, 04:16 PM

Originally Posted by razemsoft21

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

Use $cos(t+ \pi/4)= cos(t)cos(\pi/4)- sin(t)sin(\pi/4)$ $= (\sqrt{2}/2)cos(t)- (\sqrt{2}/2)sin(t)$ so $cos^2(t+ \pi/4)= \frac{1}{2}cos^2(t)- sin(t)cos(t)+ \frac{1}{2}sin^2(t)$

$\frac{cos^2(t+ \pi/4)- \frac{1}{2}}{t}= \frac{1}{2}\frac{cos^2(t)-1}{t}- \frac{sin(t)}{t}cos(t)+ \frac{1}{2}\frac{sin(t)}{t}sin(t)$ $= -\frac{1}{2}\frac{sin(t)}{t}sin(t)- \frac{sin(t)}{t}cos(t)+\frac{1}{2}\frac{sin(t)}{t} sin(t)$ $= -\frac{sin(t)}{t}cos(t)$

**razemsoft21** · July 1st 2009, 12:35 PM

Originally Posted by o_O

If you know anything about derivatives ...

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

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

Thank you for help, it works:

razemsoft21

o_O

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

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