Can you help me solve this without using de l'Hôpital's

HeescandHenv1i

HeescandHenv1i

Answered question

2022-02-26

Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):
limx0sinxxcosxx3

Answer & Explanation

Abbey Hope

Abbey Hope

Beginner2022-02-27Added 7 answers

Letting y=x3, note that your limit is the same as
limy0sin(y13)y13cos(y13)y
If you let the mean value theorem be one of your standard rules, then you can say that sin(y13)y13cos(y13)y=f(z) for some z between 0 and y. Calculating the derivative reveals that
f(z)=13sin(z13)z13
Observe that as y goes to zero, so does z, and therefore so does z13. Using that limx0sin(x)x=1 one sees therefore that
limy0f(z)=13
So 13 is the limit.
tardanetkd2

tardanetkd2

Beginner2022-02-28Added 9 answers

Use that
sin(x)=xx33!+x55!±
and that
cos(x)=1x22+x44!±
Hence
sin(x)xcos(x)x3=xx33!+x55!x+x32x54!±x3=13+x25!x24!±
Hence the Limit of
limx0sin(x)xcos(x)x3=limx013+x25!x24!=13

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