Limit of <munder> <mo form="prefix">lim <mrow class="MJX-TeXAtom-ORD"> x

flightwingsd2

flightwingsd2

Answered question

2022-06-26

Limit of lim x 0 sin ( 5 x ) sin ( 4 x )

Answer & Explanation

Stevinivm

Stevinivm

Beginner2022-06-27Added 18 answers

lim x 0 sin ( 5 x ) sin ( 4 x ) =
lim x 0 sin ( 5 x ) 5 x 4 x sin ( 4 x ) 5 x 4 x =
lim x 0 sin ( 5 x ) 4 x 5 x 5 x sin ( 4 x ) 4 x =
lim x 0 ( sin ( 5 x ) 5 x 4 x sin ( 4 x ) 5 x 4 x ) =
( lim x 0 sin ( 5 x ) 5 x ) ( lim x 0 4 x sin ( 4 x ) ) ( lim x 0 5 x 4 x ) =
( lim x 0 sin ( 5 x ) 5 x ) ( lim x 0 4 x sin ( 4 x ) ) ( lim x 0 5 4 ) =
1 1 5 4 =
5 4
crossoverman9b

crossoverman9b

Beginner2022-06-28Added 5 answers

Alternatively, you could use L'Hopital's rule:
lim x 0 sin ( 5 x ) sin ( 4 x ) = lim x 0 d d x sin ( 5 x ) d d x sin ( 4 x ) = lim x 0 5 cos ( 5 x ) 4 cos ( 4 x ) = 5 4

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?