Question

find limit if it existed

1)lim x goes to 0
tanx/2x^2-8x

2)lim x goes to-2
x^3+8/x^4-16

Answer 1

use l'Hospital's Rule:

#1 the limit is the same as

sec^2(x)/(4x-8) = -1/8

Factoring top and bottom, you have

(x+2)(x^2-2x+4) / (x-2)(x+2)(x^2+4)
So, at everywhere except x = -2,
f(x) = g(x) = (x^2-2x+4) / (x-2)(x^2+4)
g(-2) = (4+4+4)/(-4(8))
= 12/-32 = -3/8
Since f(x) just has a "hole" at x = -2, that's the limit.

Or, taking derivatives, the limit is the same as

3x^2/4x^3 = 3/4x = -3/8