We have the following limits: lim x → π 4 1 − tan 2 ⁡...

babyagelesszj

babyagelesszj

Answered

2022-07-09

We have the following limits:
lim x π 4 1 tan 2 x cos 2 x and lim θ 1 = θ 1 + sin ( θ 2 1 ) θ 2 1
Using L'Hôpital I obtained that the second limit is θ = 3 2 . For the first limit I was able to rewrite it to lim x π 4 ( 1 tan 2 ( x ) ) sec ( 2 x ), but I don't really know what to do with that.

Answer & Explanation

Marisol Morton

Marisol Morton

Expert

2022-07-10Added 13 answers

For the first limit, some elementary trigonometry (a duplication formula) will do:
1 tan 2 x c o s 2 x = cos 2 x sin 2 x cos 2 x cos 2 x = cos 2 x cos 2 x cos 2 x = 1 cos 2 x x π 4 2.
For the second limit, use asymptotic analysis:
θ 1 + sin ( θ 2 1 ) θ 2 1 = θ 1 + θ 2 1 + o ( θ 2 1 ) θ 2 1 = 1 θ + 1 + 1 + o ( 1 ) θ 1 1 2 + 1.
Jonathan Miles

Jonathan Miles

Expert

2022-07-11Added 3 answers

use the lim x 0 sin x x = 1
θ 1 + sin ( θ 2 1 ) θ 2 1 = θ 1 θ 2 1 + sin ( θ 2 1 ) θ 2 1
= 1 θ + 1 + sin ( θ 2 1 ) θ 2 1
lim x 1 ( 1 θ + 1 + sin ( θ 2 1 ) θ 2 1 ) = 1 / 2 + 1 = 3 / 2

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

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

Didn't find what you were looking for?