Question

Use series to evaluate limit below?

`\lim_(x\rarr0)(2\color(red)(\sinx)-\color(blue)(\tan^-1x)-x)/(2x^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)!)`
• `\color(blue)(\tan^-1x)=\sum_(n=0)^\infty(-1)^n(x^(2n+1))/(2n+1)`

Answer 1

`lim_(x->0)(2sin(x)-tan^-1(x)-x)/(2x^5)=-11/120`

Explanation:

We want to solve

`lim_(x->0)(2sin(x)-tan^-1(x)-x)/(2x^5)`

Using the series for `color(red)(sin(x))` and `color(blue)(tan^-1(x))`

Remember

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

`color(blue)(tan^-1(x))=sum_(n=0)^oo(-1)^n(x^(2n+1))/((2n+1))=x-x^3/3+x^5/5+...`

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`