If $L=\lim _{x \rightarrow 0} \frac{\sin x^4-x^4 \cos x^4+x^{20}}{x^4\left(e^{2 x^4}-1-2 x^4\right)}

Question

If $L=\lim _{x \rightarrow 0} \frac{\sin x^4-x^4 \cos x^4+x^{20}}{x^4\left(e^{2 x^4}-1-2 x^4\right)}$, then the value of $1 / L$ is $\_\_\_\_$

Professor PC Thomas Classes · Answer

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