# Thread: evaluating limits of trig functions #2

1. ## evaluating limits of trig functions #2

Can someone check my work for the following and fix accordingly?
Evaluate the following limits.

1. lim [1-cos^(2) x]/x^2
x->0

2. lim [1-cos2x]/x^2
x->0

3. lim cotx/(pi/2 - x)
x->0

my work:

1. lim (sin^(2)x)/x^2
x->0
i'm stuck here ..

2. lim [sin^2 (x) + cos^2 (x) - cos^ 2(x) + sin^2 (x)]/ x^2
x->0
= lim [2sin^(2) x]x^2
x->0
i'm not sure what step to take next

3. lim [cos (pi/2 - x)]/ [sin (pi/2 -x)]
(pi/2 - x) -> 0
stuck here too..

2. Originally Posted by skeske1234

1. lim (sin^(2)x)/x^2
x->0
i'm stuck here ..

Are you trying to use L'hospitals rule here? If so consider that

$\frac{d}{dx}(1-cos^2x) = sin(2x)$ and $\frac{d}{dx}(x^2) = 2x$

Now have another go at setting up the limit and consider the special case of

$\lim_{x\to 0} \frac{sin(x)}{x}=1$