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

Evaluate the following limit.
$\underset{x\to 0}{lim}\frac{\mathrm{sin}ax-{\mathrm{tan}}^{-1}ax}{b{x}^{3}}$
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The given expression is
$\underset{x\to 0}{lim}\frac{\mathrm{sin}ax-{\mathrm{tan}}^{-1}ax}{b{x}^{3}}$
Using Taylor series the expansion of sin x and ${\mathrm{tan}}^{-1}x$
$\mathrm{sin}x=\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}\frac{{x}^{2k+1}}{\left(2k+1\right)!}$
${\mathrm{tan}}^{-1}x=\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}\frac{{x}^{2k+1}}{2k+1}$
$\underset{x\to 0}{lim}\frac{\mathrm{sin}ax-{\mathrm{tan}}^{-1}ax}{b{x}^{3}}$
$=\underset{x\to 0}{lim}\frac{\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}\frac{\left(ax{\right)}^{2k+1}}{\left(2k+1\right)!}\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}\frac{\left(ax{\right)}^{2k+1}}{2k+1}}{b{x}^{3}}$
$=\underset{x\to 0}{lim}\frac{\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}\left(ax{\right)}^{2k+1}\left(\frac{1}{\left(2k+1\right)!}-\frac{1}{2k+1}\right)}{b{x}^{3}}$
$=\underset{x\to 0}{lim}\left[\frac{\left(-1{\right)}^{0}\cdot \left(a{\right)}^{2\cdot 0+1}\left(x{\right)}^{2\cdot 0-2}\left(\frac{1}{\left(2\cdot 0+1\right)!}-\frac{1}{2\cdot 0+1}\right)}{b}+\frac{\left(-1{\right)}^{1}\cdot \left(a{\right)}^{2\cdot 1+1}\left(x{\right)}^{2\cdot 1-2}\left(\frac{1}{\left(2\cdot 1+1\right)!}-\frac{1}{2\cdot 1+1}}{b}+\frac{\left(-1{\right)}^{2}\cdot \left(a{\right)}^{2\cdot 2+1}\left(x{\right)}^{2\cdot 2-2}\left(\frac{1}{\left(2\cdot 2+1\right)!}-\frac{1}{2\cdot 2+1}}{b}+...\right]$
$=\underset{x\to 0}{lim}\left[\frac{a{x}^{-2}\left(\frac{1}{1!}-\frac{1}{1}\right)}{b}-\frac{{a}^{3}{x}^{0}\left(\frac{1}{3!}-\frac{1}{3}\right)}{b}+\frac{{a}^{5}{x}^{2}\left(\frac{1}{5!}-\frac{1}{5}\right)}{b}+...\right]$
$\underset{x\to 0}{lim}\left[0-\frac{{a}^{3}\left(-\frac{1}{6}\right)}{b}+\frac{{a}^{5}{x}^{2}\left(\frac{1}{5!}-\frac{1}{5}\right)}{b}+...\right]$
$\underset{x\to 0}{lim}\left[0+\frac{{a}^{3}}{6b}+0\right]$
$=\frac{{a}^{3}}{6b}$
Jeffrey Jordon