Thursday

August 28, 2014

August 28, 2014

Posted by **claire** on Wednesday, June 18, 2014 at 7:50pm.

- math please help -
**MathMate**, Wednesday, June 18, 2014 at 10:21pmI assume you forgot the parentheses in the numerator,

Lim (sin(x)-(x^3/6))/x^5 as x->0

It would be easy to change sin(x) to a polynomial, the limits of which are easy to find.

If you expand sin(x) by Taylor's series, you'd get

sin(x)=x-x³/6+x^5/120...

So the expression becomes

(x-2x³/6+x^5/120-x^7/5040+...)/x^5 as x->0

which gives +∞

However, if the question had been

Lim (sin(x)-(x-x³/6))/x^5, the result would be 1/120

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