Derivatives and Limits

We will approach this problem by representing $f(x)$ as a power series. Using the power series formulas for the exponential and sine functions, we get $f(x)=x^2(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots)(x-\frac{x^3}{6}+\cdots).$ By carefully multiplying the two series, we get $f(x)=x^2(x-x^2-\frac{x^3}{6}+\frac{x^3}{2}+\cdots)=x^3-x^4+\frac{x^5}{6}+\cdots.$ So, $f'(x)=3x^2-4x^3+\frac{5x^4}{6}+\cdots,$ and $\lim_{x \to 0} {\frac{f'(x)}{x^2}} = \lim_{x \to 0} {\frac{3x^2-4x^3+\frac{5x^4}{6}+\cdots}{x^2}}$ $=\lim_{x \to 0} \left(3-4x+\frac{5x^2}{6}+\cdots\right) = \fbox{3}.$

f(x)=(x^2 sin⁡ x)/e^x

fʹ(x)=(e^x (x^2 cos ⁡x +2x sin x) - x^2 sin⁡ x e^x ) / e^2x

lim (x --> 0) (fʹ(x)/x^2)

= lim (x --> 0) (e^x x^2 cos ⁡x + e^x 2x sin x - x^2 sin⁡ x e^x ) / x^2 e^2x

= lim (x --> 0) (e^x x^2 cos ⁡x / x^2 e^2x ) + lim (x --> 0) ( e^x 2x sin x /x^2 e^2x ) - lim (x --> 0) ( x^2 sin⁡ x e^x /x^2 e^2x )

= lim (x --> 0) (cos⁡ x / e^x ) + lim (x --> 0) ( 2 sin x / x e^x ) - lim (x --> 0) ( sin⁡ x / e^x )

= 1 + 2 – 0

= 3