Limit exercise done without using the L'hospital Rule or series expansion asked by Falah Alnassri in the math.thi.qar group

Exercise:

Show that, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{4}}}{{{\sin }^{2}}x-{{x}^{2}}}=-3$ without using L’Hospital Rule or series expansion

Solution: Let $L=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{4}}}{{{\sin }^{2}}x-{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x.{{x}^{3}}}{\left( \sin x+x \right)\left( \sin x-x \right)}$

$\Leftrightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x+x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}}{\sin x-x}={{l}_{1}}\times {{l}_{2}}$

${{l}_{1}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x+x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{x\left( 1+\frac{\sin x}{x} \right)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{1+1}=\frac{1}{2}$

${{l}_{2}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}}{\sin x-x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{{{x}^{3}}}{{{x}^{3}}}}{\frac{\sin x-x}{{{x}^{3}}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\frac{\sin x-x}{{{x}^{3}}}}$

Remark that $\sin 3x=3\sin x-4{{\sin }^{3}}x$ hence $\sin x=3\sin \frac{x}{3}-4{{\sin }^{3}}\frac{x}{3}$

But $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x-x}{{{x}^{3}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{3\sin \frac{x}{3}-4{{\sin }^{3}}\frac{x}{3}-x}{{{x}^{3}}}$

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

Let $u=\frac{x}{3}\,\,,\,\,as\,\,\,x\to 0,u\to 0$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x-x}{{{x}^{3}}}=\frac{3}{27}\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u-u}{{{u}^{3}}}-\frac{4}{27}\underset{u\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{3}}u}{{{u}^{3}}}$

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

So ${{l}_{2}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}}{\sin x-x}=-6$ thus $L={{l}_{1}}\times {{l}_{2}}=\left( \frac{1}{2} \right)\left( -6 \right)=-3$

Therefore, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{4}}}{{{\sin }^{2}}x-{{x}^{2}}}=-3$




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The idea of solution credit to .محمد خالد غزول