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

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Answer 1 · 2018-06-26T08:38:16

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

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#