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

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

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 :

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

Taking $n=7$ then:

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

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

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

  • Am i right
  • Is there any other way

Answer

You are right. The technique you used is the best practical way to do such a problem. There are plenty of other ways, including repeated use of l’Hospital’s rule, but when you know the series representation nd there are no tricky convergence issues, that is almost always cleanest.

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