Lim x---->0 (1-cos3x)/(tan^2 3x) How can I solve this limit?

1 Answer
Jun 26, 2018

lim_(x->0) (1-cos(3x))/tan^2(3x) = 1/2

Explanation:

Divide and multiply by (3x)^2:

lim_(x->0) (1-cos(3x))/tan^2(3x) = lim_(x->0) (1-cos(3x))/(3x)^2 *(3x)^2/tan^2(3x)

Substitute t=3x. When x->0 also t->0:

lim_(x->0) (1-cos(3x))/tan^2(3x) = lim_(t->0) (1-cost)/t^2 *(t/tant)^2

Now use the well known trigonometric limits:

lim_(t->0) (1-cost)/t = 1/2

lim_(t->0) tant/t = lim_(t->0) sint/t 1/cost = 1

Then:

lim_(x->0) (1-cos(3x))/tan^2(3x) = 1/2*1^2 = 1/2