# $\lim_{x\to 0}\frac{\sin(x)-x+\frac{x^3}{3!}-\frac{x^5}{5!}}{m x^n}=\frac{8}{7!}$

If $$\lim_{x\to 0}\dfrac{\sin(x)-x+\dfrac{x^3}{3!}-\dfrac{x^5}{5!}}{m x^n}=\dfrac{8}{7!}$$

then find $m+n$:

My attempts:

note that $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + o(x^7)$$ at $0$

This makes our limit equal to :

$$\lim\limits_{x \to 0} \dfrac{- \dfrac{x^7}{7!} + o(x^7)}{mx^n}$$

Taking $n=7$ then:

$$\lim\limits_{x \to 0} \dfrac{- \dfrac{1}{7!}+o(1)}{m}$$

We can take $m=\dfrac{-1}{8}$.

then $m+n=\dfrac{-1}{8}+7$

• Am i right
• Is there any other way
• Yes it is correct. – science Feb 25 '15 at 21:04