# Find a derivative using squeeze theorem

• Feb 17th 2010, 10:30 AM
ryno16
Find a derivative using squeeze theorem
(a) State the limit definition of the derivative of a function f at the point x = a.
(b) Use (a) to find f'(0).

f(x)={x^2cos(ln|x|) if x =/= 0
{0 if x=0

I'm pretty sure you use the squeeze theorem
• Feb 17th 2010, 11:36 AM
running-gag
Hi
The cosine of any expression is always between -1 and 1
• Feb 17th 2010, 12:02 PM
vince
Quote:

Originally Posted by ryno16
(a) State the limit definition of the derivative of a function f at the point x = a.
(b) Use (a) to find f'(0).

f(x)={x^2cos(ln|x|) if x =/= 0
{0 if x=0

I'm pretty sure you use the squeeze theorem

The limit definition of $f^{'}(x)$ is
$f^{'}(x)=\lim_{h \rightarrow 0}\Bigg{(} \frac{(x+h)^{2}\cos(ln|x+h|)-x^2\cos(ln|x|)}{h} \Bigg{)}$

Now for a=0, this means

$f^{'}(0)=\lim_{h \rightarrow 0} \Bigg{(} \frac{h^2\cos(ln|h|)}{h} \Bigg{)}$ (*)

But since $-1<\cos(x)<1 , \forall x$
$\lim_{h \rightarrow 0} -h < (*) < \lim_{h \rightarrow 0} h$

Apply squeeze theorem to conclude that f('0) = 0.