Question

$\displaystyle \lim_{x \to 0} \frac{sin \,x^{4} - x^{4} \, cos \,x^{4} + x^{20}}{x^{4}\left(e^{2x^4 1-2x^4}\right)}$ is equal to

• A. 0
• B. - 1/6
• C. 1/6
• D. does not exist

Answer 1

Answer: (C)

$\displaystyle \lim_{x \to 0} \frac{sin \,x^{4} - x^{4} \, cos \,x^{4} + x^{20}}{x^{4}\left(e^{2x^4 1-2x^4}\right)}$
$= \displaystyle \lim_{t \rightarrow 0} \frac{sin \,t-t cos \,t +t^{5}}{t\left(e^{2t}-1-2t\right)}$
$= \displaystyle \lim_{t \rightarrow 0} \frac{t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}..... -t \left(1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}..........\right)+t^{5}}{t\left(1+2t+ \frac{4t^{2}}{2!}+\frac{8t^{3}}{3!}+\frac{16t^{4}}{4!}+....-1-2t\right)}$
$= \displaystyle \lim_{t \rightarrow 0} \frac{-\frac{t^{3}}{6}+\frac{t^{3}}{2}+\frac{t^{5}}{5!}-\frac{t^{5}}{4!}+..... +t^{5}}{2t^{3}+\frac{8t^{4}}{3!}+.......}$
$= \frac{-\frac{1}{6}+\frac{1}{2}}{2} =-\frac{-1+3}{12} = \frac{1}{6}$