Question

# Evaluate the following limit. lim_{xrightarrow0}frac{sin ax-tan^{-1}ax}{bx^3}

Limits and continuity
Evaluate the following limit.
$$\lim_{x\rightarrow0}\frac{\sin ax-\tan^{-1}ax}{bx^3}$$

2021-01-14
The given expression is
$$\lim_{x\rightarrow0}\frac{\sin ax-\tan^{-1}ax}{bx^3}$$
Using Taylor series the expansion of sin x and $$\tan^{-1}x$$
$$\sin x=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}$$
$$\tan^{-1}x=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{2k+1}$$
$$\lim_{x\rightarrow0}\frac{\sin ax-\tan^{-1}ax}{bx^3}$$
$$=\lim_{x\rightarrow0}\frac{\sum_{k=0}^\infty(-1)^k\frac{(ax)^{2k+1}}{(2k+1)!}\sum_{k=0}^\infty(-1)^k\frac{(ax)^{2k+1}}{2k+1}}{bx^3}$$
$$=\lim_{x\rightarrow0}\frac{\sum_{k=0}^\infty(-1)^k{(ax)^{2k+1}}(\frac{1}{(2k+1)!}-\frac{1}{2k+1})}{bx^3}$$
$$=\lim_{x\rightarrow0}[\frac{(-1)^0\cdot(a)^{2\cdot0+1}(x)^{2\cdot0-2}(\frac{1}{(2\cdot0+1)!}-\frac{1}{2\cdot0+1})}{b}+\frac{(-1)^1\cdot(a)^{2\cdot1+1}(x)^{2\cdot1-2}(\frac{1}{(2\cdot1+1)!}-\frac{1}{2\cdot1+1}}{b}+\frac{(-1)^2\cdot(a)^{2\cdot2+1}(x)^{2\cdot2-2}(\frac{1}{(2\cdot2+1)!}-\frac{1}{2\cdot2+1}}{b}+...]$$
$$=\lim_{x\rightarrow0}[\frac{ax^{-2}(\frac{1}{1!}-\frac{1}{1})}{b}-\frac{a^3x^0(\frac{1}{3!}-\frac{1}{3})}{b}+\frac{a^5x^2(\frac{1}{5!}-\frac{1}{5})}{b}+...]$$
$$\lim_{x\rightarrow0}[0-\frac{a^3(-\frac{1}{6})}{b}+\frac{a^5x^2(\frac{1}{5!}-\frac{1}{5})}{b}+...]$$
$$\lim_{x\rightarrow0}[0+\frac{a^3}{6b}+0]$$
$$=\frac{a^3}{6b}$$