Simplifying ln to solve L'Hôpital's rule <munder> <mo movablelimits="true" form="prefix">lim

hawwend8u

hawwend8u

Answered question

2022-05-22

Simplifying ln to solve L'Hôpital's rule
lim x 0.5 [ ( 2 x 1 ) 4 ln ( 1 2 x ) ]

Answer & Explanation

Ronnie Glenn

Ronnie Glenn

Beginner2022-05-23Added 11 answers

Write it as
ln ( 1 2 x ) ( 2 x 1 ) 4
This is an form. Differentiating the top and the bottom (by the chain rule for example), we get
2 ( 1 2 x ) 1 8 ( 2 x 1 ) 5 = 1 4 ( 2 x 1 ) 5 1 2 x = 1 4 ( 2 x 1 ) 4
Now lim x 0.5 1 4 ( 2 x 1 ) 4 = 0
America Ware

America Ware

Beginner2022-05-24Added 1 answers

Let y = 2 x 1 x = y + 1 2
lim x 0.5 ( 2 x 1 ) 4 ln ( 1 2 x )
= lim y 0 y 4 ln ( y )
= lim y 0 ln ( y ) 1 y 4
= lim y 0 1 / y 4 / y 5
= lim y 0 1 / y 4 / y 5
= lim y 0 y 4 4
= 0

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