Question

$\underset{x\to a}{\lim }\frac{a\sin x-x\sin a}{a{x}^2-x{a}^2}$

Answer 1

$$\begin{gathered} \underset{x\to a}{\lim }\left[\frac{a\sin x-x\sin a}{a{x}^2-x{a}^2}\right] \\ \mathrm{Let}x=a+h \\ \mathrm{When}x\to a,\mathrm{then}h\to 0. \\ =\underset{h\to 0}{\lim }\left[\frac{a\sin \left(a+h\right)-\left(a+h\right)\sin a}{a{\left(a+h\right)}^2-\left(a+h\right)a^2}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{a\sin \left(a+h\right)-a\sin a-h\sin a}{a\left(a^2+h^2+2ah\right)-a^3-a^{2}h}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{a\left[\sin \left(a+h\right)-\sin a\right]-h\sin a}{a^3+a{h}^2+2a^{2}h-a^3-a^{2}h}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{a\times 2\cos \left({\displaystyle \frac{a+h+a}{2}}\right)\sin \left({\displaystyle \frac{a+h-a}{2}}\right)-h\sin a}{a{h}^2+a^{2}h}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{a\times 2\cos \left({\displaystyle \frac{2a+h}{2}}\right)\sin \left({\displaystyle \frac{h}{2}}\right)-h\sin a}{ah\left(h+a\right)}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{2a\cos \left({\displaystyle \frac{2a+h}{2}}\right)\sin \left({\displaystyle \frac{h}{2}}\right)}{a\left(a+h\right)\times {\displaystyle \frac{h}{2}}\times 2}-\frac{h\sin a}{ah\left(a+h\right)}\right] \\ =\frac{2a\cos a\times 1}{a^2\times 2}-\frac{\sin a}{a^2} \\ =\frac{a\cos a-\sin a}{a^2} \end{gathered}$$