I need to find <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXA

Kaeden Woodard

Kaeden Woodard

Answered question

2022-05-19

I need to find
lim x 0 e x sin x x ( 1 x ) x 3

Answer & Explanation

thoumToofwj

thoumToofwj

Beginner2022-05-20Added 16 answers

Actually,
lim x 0 + e x cos ( x ) + 1 3 x = ,
and it follows from L'Hopital's Rule that
lim x 0 + e x sin ( x ) x ( 1 x ) x 3 = .
By the same argument,
lim x 0 e x sin ( x ) x ( 1 x ) x 3 = .
So, the limit
lim x 0 e x sin ( x ) x ( 1 x ) x 3
lim x 0 e x sin ( x ) x ( 1 x ) x 3
doesn't exist indeed.
Alaina Marshall

Alaina Marshall

Beginner2022-05-21Added 5 answers

As an alternative way, we can avoid l'Hopital's rule noting that
e x sin x x ( 1 x ) x 3 = e x sin x sin x + sin x x ( 1 x ) x 3 =
= e x sin x sin x x 3 + sin x x ( 1 x ) x 3 =
= 1 x sin x x e x 1 x + sin x x + x 2 x 3 =
= 1 x sin x x e x 1 x + sin x x x 3 + 1 x
with sin x x 1, e x 1 x 1, sin x x x 3 1 6 but limit for 1 x doesn't exist.

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