I am having trouble finding the limit for the following function: <munder> <mo movablelimits

EnvivyEvoxys6

EnvivyEvoxys6

Answered question

2022-07-08

I am having trouble finding the limit for the following function:
lim x 0 f ( x ) = tan 3 x tan 2 x
I do not want to use L'Hospital's rule.
I know that tan 3 x tan 2 x = sin 3 x cos 2 x sin 2 x cos 3 x , but that does not help me.
Can you maybe give me a hint?

Answer & Explanation

Caiden Barrett

Caiden Barrett

Beginner2022-07-09Added 20 answers

sin 3 x cos 2 x sin 2 x cos 3 x = 3 2 . sin 3 x 3 x . 2 x sin 2 x . cos 2 x cos 3 x
pipantasi4

pipantasi4

Beginner2022-07-10Added 6 answers

Using tan ( A + B ) = tan A + tan B 1 tan A tan B twice we obtain
tan 2 x = 2 tan x 1 tan 2 x ; tan 3 x = 3 tan x tan 3 x 1 3 tan 2 x
so that
tan 2 x tan 3 x = 2 ( 1 3 tan 2 x ) ( 3 tan 2 x ) ( 1 tan 2 x )
And (having cancelled a factor of tanx) nothing vanishes and the limit is easy to find, even if you know no other limits.

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?