# Thread: Find the limit of a function

1. ## Find the limit of a function

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

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

3. 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 $f(x) = \cos^2 (x)$

4. Originally Posted by Krizalid
just put $t=x-\frac\pi4.$
OK , I put $t=x-\frac\pi4$
I got the following restul !!

5. 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 $\frac{sinx}{x}$

6. 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 ....

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

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

8. ## Thanks

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