Continuity exercise without using L'hopital Rule ,series and any advanced Math analysis

Exercise:

Let $f,g$ be two functions to be defined as follows:

\[f\left( x \right)=\frac{x-\sin x}{{{x}^{3}}}\,\,\,\And \,\,g\left( x \right)=\frac{{{e}^{x}}-{{e}^{-x}}-2x}{x-\sin x}\]

1. Show that , $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\frac{1}{6}$
2. Deduce that , $\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=2$

Solution:

1. 
   

Let $x=3z$ so $f\left( 3z \right)=\frac{3z-\sin \left( 3z \right)}{27{{z}^{3}}}$

but $\sin 3z=3{{\cos }^{2}}z\sin z-{{\sin }^{3}}z=3\left( 1-{{\sin }^{2}}z \right)\sin z-{{\sin }^{3}}z$

$\Rightarrow \sin 3z=3\sin z-4{{\sin }^{3}}z$

Thus $f\left( 3z \right)=\frac{3z-3\sin z+4{{\sin }^{3}}z}{27{{z}^{3}}}$  as $x\to 0\,\,,\,\,z\to 0$

So $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{z\to 0}{\mathop{\lim }}\,f\left( 3z \right)=\underset{z\to 0}{\mathop{\lim }}\,\frac{3z-3\sin z+4{{\sin }^{3}}z}{27{{z}^{3}}}=\underset{z\to 0}{\mathop{\lim }}\,\frac{3\left( z-\sin z \right)}{27{{z}^{3}}}+\underset{z\to 0}{\mathop{\lim }}\,\frac{4{{\sin }^{3}}z}{27{{z}^{3}}}$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin x}{{{x}^{3}}}=\frac{1}{9}\underset{z\to 0}{\mathop{\lim }}\,\frac{z-\sin z}{{{z}^{3}}}+\frac{4}{27}\underset{z\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{3}}z}{{{z}^{3}}}$  but $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1$

Thus $\left( 1-\frac{1}{9} \right)\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin x}{{{x}^{3}}}=\frac{4}{27}\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin x}{{{x}^{3}}}=\frac{9}{8}\times \frac{4}{27}=\frac{1}{6}$

   2.

We have $g\left( x \right)=\frac{{{e}^{x}}-{{e}^{-x}}-2x}{x-\sin x}=\frac{2\left( \frac{{{e}^{x}}-{{e}^{-x}}}{2}-x \right)}{x-\sin x}=\frac{2\left( \sinh x-x \right)}{x-\sin x}=\frac{\frac{2\sinh x-2x}{{{x}^{3}}}}{\frac{x-\sin x}{{{x}^{3}}}}$

So \(\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{2\sinh x-2x}{{{x}^{3}}}}{\frac{x-\sin x}{{{x}^{3}}}}=\frac{\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sinh x-2x}{{{x}^{3}}}}{\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin x}{{{x}^{3}}}}=\frac{\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sinh x-2x}{{{x}^{3}}}}{\frac{1}{6}}\)

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{12\left( \sinh x-x \right)}{{{x}^{3}}}=12\underset{x\to 0}{\mathop{\lim }}\,\frac{\sinh x-x}{{{x}^{3}}}$ Take $x=3w$ as $x\to 0\,\,,\,\,w\to 0\,$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sinh x-x}{{{x}^{3}}}=\underset{w\to 0}{\mathop{\lim }}\,\frac{\sinh \left( 3w \right)-3w}{{{\left( 3w \right)}^{3}}}=\underset{w\to 0}{\mathop{\lim }}\,\frac{\sinh \left( 3w \right)-3w}{27{{w}^{3}}}$

But $\sinh \left( 3w \right)=3{{\cosh }^{2}}w\sinh w+{{\sinh }^{3}}w=3\sinh w+4{{\sinh }^{3}}w$

So $\underset{w\to 0}{\mathop{\lim }}\,\frac{\sinh \left( 3w \right)-3w}{27{{w}^{3}}}=\underset{w\to 0}{\mathop{\lim }}\,\frac{3\sinh w+4{{\sinh }^{3}}w-3w}{27{{w}^{3}}}$

$=\underset{w\to 0}{\mathop{\lim }}\,\frac{3\sinh w-3w}{27{{w}^{3}}}+\underset{w\to 0}{\mathop{\lim }}\,\frac{4{{\sinh }^{3}}w}{27{{w}^{3}}}$   

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sinh x-x}{{{x}^{3}}}=\frac{1}{9}\underset{w\to 0}{\mathop{\lim }}\,\frac{\sinh w-w}{{{w}^{3}}}+\frac{4}{27}\underset{w\to 0}{\mathop{\lim }}\,\frac{{{\sinh }^{3}}w}{{{w}^{3}}}$ but $\underset{w\to 0}{\mathop{\lim }}\,\frac{\sinh w}{w}=1$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sinh x-x}{{{x}^{3}}}=\frac{4}{27}\times \frac{9}{8}=\frac{1}{6}$

Thus $\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=12\underset{x\to 0}{\mathop{\lim }}\,\frac{\sinh x-x}{{{x}^{3}}}=\frac{12}{6}=2$