Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):limx→0sin⁡x−xcos⁡xx3

Use thatsin⁡(x)=x−x33!+x55!±…and thatcos⁡(x)=1−x22+x44!±…Hencesin⁡(x)−xcos⁡(x)x3=x−x33!+x55!−x+x32−x54!±…x3=13+x25!−x24!±…Hence the Limit oflimx→0sin⁡(x)−xcos⁡(x)x3=limx→013+x25!−x24!=13

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HeescandHenv1i

Answered question

2022-02-26

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

lim_{x \to 0} \frac{\sin x - x \cos x}{x^{3}}

Answer & Explanation

Abbey Hope

Beginner2022-02-27Added 7 answers

Letting

y = x^{3}

, note that your limit is the same as

lim_{y \to 0} \frac{\sin (y^{\frac{1}{3}}) - y^{\frac{1}{3}} \cos (y^{\frac{1}{3}})}{y}

If you let the mean value theorem be one of your standard rules, then you can say that

\frac{\sin (y^{\frac{1}{3}}) - y^{\frac{1}{3}} \cos (y^{\frac{1}{3}})}{y} = f^{'} (z)

for some z between 0 and y. Calculating the derivative reveals that

f^{'} (z) = \frac{1}{3} \frac{\sin (z^{\frac{1}{3}})}{z^{\frac{1}{3}}}

Observe that as y goes to zero, so does z, and therefore so does

z^{\frac{1}{3}}

. Using that

lim_{x \to 0} \frac{\sin (x)}{x} = 1

one sees therefore that

lim_{y \to 0} f^{'} (z) = \frac{1}{3}

\frac{1}{3}

is the limit.

tardanetkd2

Beginner2022-02-28Added 9 answers

Use that

\sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} \pm \dots

and that

\cos (x) = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} \pm \dots

Hence

\frac{\sin (x) - x \cos (x)}{x^{3}} = \frac{x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - x + \frac{x^{3}}{2} - \frac{x^{5}}{4!} \pm \dots}{x^{3}} = \frac{1}{3} + \frac{x^{2}}{5!} - \frac{x^{2}}{4!} \pm \dots

Hence the Limit of

lim_{x \to 0} \frac{\sin (x) - x \cos (x)}{x^{3}} = lim_{x \to 0} \frac{1}{3} + \frac{x^{2}}{5!} - \frac{x^{2}}{4!} = \frac{1}{3}

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