lim (x -> 0) x^2(1 + cot^2(3x))?

Question

How do I evaluate this?

Thank you for the help.

Answer 1

1 + cot^2 (3x) = cosec^ 3x
thus your ratio is the same as
x^2/ sin^2 3x
and the limit is 1/9

Answer 2

Since cot(0) = cos(0)/sin(0) = 1/0 is undefined, you'll need to transform things into a ratio to be able to use l'Hospital's Rule.

Use the identity 1 + cot² = csc² to get...
lim (x -> 0) x² * sec²(3x)
= lim (x -> 0) x² / sin²(3x)
= 0/0

Differentiate top and bottom...
= lim(x→0) 2x / [6sin(3x)cos(3x)]
= lim(x→0) 2x / 3sin(6x)
= 0/0

Apply l'Hospital's Rule again...
= lim(x→0) 2 / 18cos(6x)
= 2/18
= 1/9.

Answer 3

For every x where such expression is defined, (x^2)(cot^2(3x)) = (x/ tan(3x))^2 . We know lim (x --> 0) tan(x)/x = 1. Therefore, lim (x --> 0) (x^2)(cot^2(3x)) = lim (x --> 0)(x/ tan(3x))^2 = lim (x --> 0)(1/9) (3x/ tan(3x))^2 = 1/9 * (1) = 1/9.

So, lim (x -->0) x^2(1 + cot^2(3x)) = lim (x --> 0) x^2 + lim(x -->0) x^2)(cot^2(3x)) = 0 + 1/9 = 1/9.

Answer 4

To evaluate a lim, you plug c (whatever x approaches) into the equation. If you get an undefined answer, use L'Hopital's Rule. :)

0^2(1 + cot^2(3*0)) =

0(1+cot^2(0)) = 0

That's fine, but if you've gotten farther into calc, you'll want to use L'Hopital's.

EDIT: You just gave me a thumbs down because I'm Jewish.