Limit problem

1. May 8, 2007

reza

Lim cot x^2 - 1/x^2
x--->&

2. May 8, 2007

HallsofIvy

Staff Emeritus
If you want help you must at least explain what you are doing! What is that "&"?? Do you mean cot(x^2)- (1/x^2) or (cot(x^2)- 1)/x^2 or cot(x^2-1)/x^2 or cot((x^2-1)/x^2)?

3. May 8, 2007

reza

sorry for my bad method writing
lim [(cot x^2) - (1/(x^2))]
x-->1/0

1/0 means infinite i dont know how to show its sign
sorry again

4. May 9, 2007

VietDao29

Ok, I assume that you mean:
$$\lim_{x \rightarrow \infty} \left( \cot (x ^ 2) - \frac{1}{x ^ 2} \right)$$
Do you mean that?

If that's the limit you want to take, then, you should notice that, when x tends to infinity, 1/x2 will tend to 0, right? Then, the limit will only depend on the limit of cot(x2) as x tends to infinity.

What is the limit of: $$\lim_{x \rightarrow \infty} \cot (x ^ 2)$$? Will it tend to a fixed value? Or will it diverge?

5. May 12, 2007

reza

thanks
then the answre would infinit yes?
and what wold happen if x->0

6. May 12, 2007

VietDao29

No, it's not infinity, the limit does not exist.
You should notice that:
$$\cot \left( \frac{\pi}{2} + k \pi \right) = 0 , \ \ k \in \mathbb{Z}$$
and
$$\cot \left( \frac{\pi}{4} + k \pi \right) = 1 , \ \ k \in \mathbb{Z}$$
i.e, when x tends to infinity, it does not tends to a fixed value, it varies. So the limit does not exist.

Do you mean:
$$\lim_{x \rightarrow 0} \left( \cot ^ 2 x - \frac{1}{x ^ 2} \right)$$?
Have you cover L'Hopital rule? What have you tried? :)

7. May 13, 2007

reza

thanks
then sin cos cot and tan when x tends to infinity the limit does not exist

__________________________

lim (cot^2 x-(1/x^2))=lim [(cos^2 x -1)/x^2]=lim [(-2sinxcosx)/2x] =
x->0 x->0 x->0
=lim (-sin2x/2x) =lim (-2cosx/2)=-1/2
x->0 x->0
is it right?

8. May 13, 2007

HallsofIvy

Staff Emeritus
How do you get from cot2 x- (1/x2) to [cos2x- 1]/x2?

9. May 14, 2007

reza

excuse me i put mistakaly sinx^2~x^2

lim (cot^2 x-(1/x^2))=lim [(1/tan^2 x) - (1/x^2)]=..
and set 1/tan x^2 = (1/sin x^2)-1
is it right?