[hrcs]

Could anyone help me find the

limit as x->0 of (1/x^2)-(cotx)^2 without the use of a calculator?

I tried using L'Hospitals rule but the result is too messy to work with. Any help or thoughts would be appreciated.

[gamehead200]

Have you expanded it to:

lim x-> 0
(x^-2)-((cos(x))^2)/((sin(x))^2)

Just a thought. Seems simpler this way.

[LLXX]

Common denominator and trigonometric identities.

This post has been edited by LLXX: 08 December 2006 - 02:45 AM

[Innocent Devil]

try Lt x->0 (1- x^2.cot(x)^2)/x^2

and apply L'Hospitals Rule

withing 2rd iteration
u get Denominator as 2 and then u have to limit only the Nr.

It lead to either Infinity or Zero

[ripken204]

well the 1/x^2 part is equal to 0

the cot part is just insane without a calc tho. but it would be -.025

[gamehead200]

ripken204, on Dec 8 2006, 11:26 PM, said:

well the 1/x^2 part is equal to 0

the cot part is just insane without a calc tho. but it would be -.025

Why not just make it cos^2(x)/sin^2(x)?

And the limit is equal to 2/3, actually. Just used my TI-89 to solve it.

[allen2]

Here is my solution:
(1/x^2)-(cotx)^2= 1/x^2 - (cos(x)/sin(x))^2
= 1/x^2- (cos(x)^2/sin(x)^2)
= 1/x^2 - (1 -sin(x)^2)/sin(x)^2 cos(x)^2 +sin(x)^2 =1
=1/x^2 - (1/sin(x)^2 -1)
= 1/x^2 - 1/sin(x)^2 +1
= (sin(x)^2 -x^2)/x^2 sin(x)^2 +1
with MacLaurin series for sin(x)= x-x^3/3! when x->0
=( (x-x^3/3!)^2-x^2)/(x^2(x-x^3/3!)^2)+1
=(x^2 -2x^4/3! +x^6/36 -x^2)/(x^2 (x(1-x^2/3!))^2) +1
=( -x^4/3+x^6/36)/(x^2*x^2(1-x^2/3!)^2) +1
=(x^4 (-1/3 +x^2/36))/(x^4 (1-x^2/6)^2) +1
=(-1/3 +x^2/36)/(1-x^2/6)^2 +1
And when x->0 this (-1/3 +x^2/36)/(1-x^2/6)^2 -> -1/3
So Lim x->0 of (1/x^2)-(cotx)^2= -1/3+1=2/3

[hrcs]

allen2, on Dec 9 2006, 06:02 AM, said:

Here is my solution:
(1/x^2)-(cotx)^2= 1/x^2 - (cos(x)/sin(x))^2
= 1/x^2- (cos(x)^2/sin(x)^2)
= 1/x^2 - (1 -sin(x)^2)/sin(x)^2 cos(x)^2 +sin(x)^2 =1
=1/x^2 - (1/sin(x)^2 -1)
= 1/x^2 - 1/sin(x)^2 +1
= (sin(x)^2 -x^2)/x^2 sin(x)^2 +1
with MacLaurin series for sin(x)= x-x^3/3! when x->0
=( (x-x^3/3!)^2-x^2)/(x^2(x-x^3/3!)^2)+1
=(x^2 -2x^4/3! +x^6/36 -x^2)/(x^2 (x(1-x^2/3!))^2) +1
=( -x^4/3+x^6/36)/(x^2*x^2(1-x^2/3!)^2) +1
=(x^4 (-1/3 +x^2/36))/(x^4 (1-x^2/6)^2) +1
=(-1/3 +x^2/36)/(1-x^2/6)^2 +1
And when x->0 this (-1/3 +x^2/36)/(1-x^2/6)^2 -> -1/3
So Lim x->0 of (1/x^2)-(cotx)^2= -1/3+1=2/3

Thanks.. though we're technically not supposed to know the MacLaurin or Taylor series right now