# Use series to evaluate limit below?

## $\setminus {\lim}_{x \setminus \rightarrow 0} \frac{2 \setminus \textcolor{red}{\setminus \sin x} - \setminus \textcolor{b l u e}{\setminus {\tan}^{-} 1 x} - x}{2 {x}^{5}}$ I know the following MacLaurin formulas (which might be applicable here): \color(red)\sinx=\sum_(n=0)^\infty(-1)^n(x^(2n+1))/((2n+1)!) $\setminus \textcolor{b l u e}{\setminus {\tan}^{-} 1 x} = \setminus {\sum}_{n = 0}^{\setminus} \infty {\left(- 1\right)}^{n} \frac{{x}^{2 n + 1}}{2 n + 1}$

May 27, 2018

${\lim}_{x \to 0} \frac{2 \sin \left(x\right) - {\tan}^{-} 1 \left(x\right) - x}{2 {x}^{5}} = - \frac{11}{120}$

#### Explanation:

We want to solve

${\lim}_{x \to 0} \frac{2 \sin \left(x\right) - {\tan}^{-} 1 \left(x\right) - x}{2 {x}^{5}}$

Using the series for $\textcolor{red}{\sin \left(x\right)}$ and $\textcolor{b l u e}{{\tan}^{-} 1 \left(x\right)}$

Remember

color(red)(sin(x))=sum_(n=0)^oo(-1)^n(x^(2n+1))/((2n+1)!)=x-x^3/(3!)+x^5/(5!)+...

$\textcolor{b l u e}{{\tan}^{-} 1 \left(x\right)} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{{x}^{2 n + 1}}{\left(2 n + 1\right)} = x - {x}^{3} / 3 + {x}^{5} / 5 + \ldots$

Thus

lim_(x->0)(2color(red)(sum_(n=0)^oo(-1)^n(x^(2n+1))/((2n+1)!))-color(blue)(sum_(n=0)^oo(-1)^n(x^(2n+1))/((2n+1)))-x)/(2x^5)

=lim_(x->0)(2(color(red)(x-x^3/(3!)+x^5/(5!)...))-(color(blue)(x-x^3/3+x^5/5+...))-x)/(2x^5)

But, we only need a few terms of the series
Now i cut off some of the series, because the next terms of the series, will have a higher power than the numerator does and therefore will approach 0, as x approaches 0

lim_(x->0)(2(x-x^3/(3!)+x^5/(5!))-(x-x^3/3+x^5/5)-x)/(2x^5)

Remove the brackets

lim_(x->0)(2x-2x^3/(3!)+2x^5/(5!)-x+x^3/3-x^5/5-x)/(2x^5)

By cancelling out

lim_(x->0)(2x^5/(5!)-x^5/5)/(2x^5)=lim_(x->0)(2/(5!)-1/5)/2=-11/120