find the value of : lim tan(x) / ( 1 - cos(x) ) when x-->0?

Question

find the value of :
lim tan(x) / ( 1 - cos(x) )
x-->0

Answer 1

Multiply x/x
x tan x / x (1-cos x)
(sin x / cos x )x * (1/ 1- cos x)x
Rearrange
(sin x /x cos x )* (1x/ 1- cos x)
(sin x /x )(1/cos x )* (1x/ 1- cos x)
x->0 (sin x /x) = 1
1/cos x = 1
1x/ 1- cos x = reciprocal (1-cos x)/x
= -(cos x - 1)/x
Since lim x->0 for (cos x-1)/x = 1
So -(cos x - 1)/x = -1
(sin x /x )*(1/cos x )* (x/ 1- cos x)
1 * 1 * -1 = -1
The limit is -1.

Answer 2

First tan x = sin x/cos x
so the expression becomes
lim(x→0) sin x/(1 - cos x) * 1/cos x.
As x →0, the second term goes to 1,
so we have to evaluate
lim(x→0) sin x/ (1- cos x).
Now multiply numerator and denominator by
1 + cos x and use 1- cos² x = sin² x. We get
lim(x →0) sin x(1 + cos x)/ sin² x =
lim(x→0) (1+cos x)/sin x, which is infinity.

Answer 3

Since substituting 0 for x yields:
0/(1 - 1) = 0/0, or an undefined quotient, apply L'Hospital's Rule which yields:
sec^2x/[-(-sin(x)]. Now substituting in 0 yields: 1 /[0], or infinity!