Question

Show that as x->0, sinx = x - x^3/3! + o(x^4)

Answer 1

that is the Taylor expansion of the function sin x

Answer 2

what is taylor expansion?

Answer 3

It is a power series, here are the first three terms of the Taylor expansion:
\[\Large \begin{gathered}
\sin x = \sin \left( 0 \right) + {\left. {\frac{{d\sin x}}{{dx}}} \right|_{x = 0}}x + \frac{1}{{2!}}{\left. {\frac{{{d^2}\sin x}}{{d{x^2}}}} \right|_{x = 0}}{x^2} + \hfill \\
\hfill \\
+ {\left. {\frac{1}{{3!}}\frac{{{d^3}\sin x}}{{d{x^3}}}} \right|_{x = 0}}{x^3} \hfill \\
\end{gathered} \]

Answer 4

for example, we have:
\[\Large {\left. {\frac{{d\sin x}}{{dx}}} \right|_{x = 0}} = \cos \left( 0 \right) = 1\]

Answer 5

so would it be, sinx = x- x^3/3! and then because limit x->0 (sinx-x-x^3/3!) = lim x->( sinx-sinx) = 0, then sinx = x-x^3/x! + o(x^4)?