Evaluate <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-

Palmosigx

Palmosigx

Answered question

2022-07-12

Evaluate lim x 0 tan ( tan x ) sin ( sin x ) tan x sin x

Answer & Explanation

sniokd

sniokd

Beginner2022-07-13Added 22 answers

lim x 0 tan ( tan x ) sin ( sin x ) tan x sin x = lim x 0 tan ( tan x ) tan ( sin x ) cos ( sin x ) tan x sin x = lim x 0 tan ( tan x ) tan ( sin x ) + tan ( sin x ) tan ( sin x ) cos ( sin x ) tan x sin x = lim x 0 ( tan ( tan x ) tan ( sin x ) tan x sin x + tan ( sin x ) ( 1 cos ( sin x ) ) tan x sin x )
Then, by mean value theorem, there is c ( π 2 , π 2 ) such that
tan ( tan x ) tan ( sin x ) tan x sin x = sec 2 c
and between sin x and tan x. Then lim x 0 tan ( tan x ) tan ( sin x ) tan x sin x = 1. Next, we will show that lim x 0 tan ( sin x ) ( 1 cos ( sin x ) ) tan x sin x = 1
lim x 0 tan ( sin x ) ( 1 cos ( sin x ) ) tan x sin x = lim x 0 tan ( sin x ) ( 1 cos ( sin x ) ) cos x sin x ( 1 cos x ) = lim x 0 tan ( sin x ) sin x 1 cos ( sin x ) sin 2 x sin 2 x 1 cos x cos x = lim x 0 tan ( sin x ) sin x lim x 0 1 cos ( sin x ) sin 2 x lim x 0 ( 1 + cos x ) lim x 0 cos x = 1 1 2 2 1 = 1.
Therefore,
lim x 0 tan ( tan x ) sin ( sin x ) tan x sin x = lim x 0 tan ( tan x ) tan ( sin x ) tan x sin x + lim x 0 tan ( sin x ) ( 1 cos ( sin x ) ) tan x sin x = 1 + 1 = 2

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