Finding Limits Using the Cotangent and Tangent Functions

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Homework Help Overview

The discussion revolves around finding limits involving the cotangent and tangent functions, specifically the expression lim (cot(x^2) - 1/x^2) as x approaches various values, including infinity and zero.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the interpretation of the limit expression and clarify the notation used. Questions arise regarding the behavior of cotangent as x approaches infinity and zero, as well as the implications of the limit's existence or divergence.

Discussion Status

Some participants have offered insights into the behavior of cotangent at infinity, noting that it does not converge to a fixed value. Others are questioning the steps taken in the limit calculations and the application of L'Hôpital's rule, indicating a mix of understanding and confusion about the problem.

Contextual Notes

There are indications of miscommunication regarding the limit notation and the assumptions about the behavior of trigonometric functions at specific points, particularly concerning infinity and zero.

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

1/0 means infinite i don't know how to show its sign
sorry again
 
Ok, I assume that you mean:
[tex]\lim_{x \rightarrow \infty} \left( \cot (x ^ 2) - \frac{1}{x ^ 2} \right)[/tex]
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: [tex]\lim_{x \rightarrow \infty} \cot (x ^ 2)[/tex]? Will it tend to a fixed value? Or will it diverge?
 
thanks
then the answre would infinit yes?
and what wold happen if x->0
 
reza said:
thanks
then the answre would infinit yes?

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

and what wold happen if x->0

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

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

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