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

Averi Mitchell

Averi Mitchell

Answered question

2022-06-25

lim x ( x 6 [ e 1 2 x 3 cos ( 1 x x ) ] )
Any tip on how to calculate it?

Answer & Explanation

ejigaboo8y

ejigaboo8y

Beginner2022-06-26Added 29 answers

In such cases, I always find that transforming the beast into a limit at zero is easier. Perform the substitution
x 3 / 2 = 1 / t
so x 3 = 1 / t 2 and x 6 = 1 / t 4 . Now the limit becomes
lim t 0 + e t 2 / 2 cos t t 4
and the Taylor expansion of the numerator up to degree 4 is
1 t 2 / 2 1 ! + t 4 / 4 2 ! 1 + t 2 2 ! t 4 4 ! + o ( t 4 ) = t 4 12 + o ( t 4 )

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