Limit exercise without using L'Hoppital rule or series asked by katrina linda in pretty ath group

Exercise :

Show that, $\underset{x\to 0}{\mathop{\lim }}\,\,\left( \,\frac{1}{{{\sin }^{2}}x}-\frac{1}{{{x}^{2}}} \right)=\frac{1}{3}$

Solution: Let $x=2t$ so as $x\to 0\,\Leftrightarrow t\to 0$

So $L=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{\sin }^{2}}x}-\frac{1}{{{x}^{2}}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{1}{{{\sin }^{2}}2t}-\frac{1}{4{{t}^{2}}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{1}{{{\left( 2\sin t\cos t \right)}^{2}}}-\frac{1}{4{{t}^{2}}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{1}{4{{\sin }^{2}}t{{\cos }^{2}}t}-\frac{1}{4{{t}^{2}}}$

Thus $4L=\underset{t\to 0}{\mathop{\lim }}\,\frac{1}{{{\sin }^{2}}{{\cos }^{2}}t}-\frac{1}{{{t}^{2}}}+L-L$

$\Rightarrow 3L=\underset{t\to 0}{\mathop{\lim }}\,\frac{1}{{{\sin }^{2}}t{{\cos }^{2}}t}-\frac{1}{{{t}^{2}}}+L$

So $3L=\underset{t\to 0}{\mathop{\lim }}\,\frac{1}{{{\sin }^{2}}t{{\cos }^{2}}t}-\frac{1}{{{t}^{2}}}-\underset{t\to 0}{\mathop{\lim }}\,\left( \frac{1}{{{\sin }^{2}}t}-\frac{1}{{{t}^{2}}} \right)=\underset{t\to 0}{\mathop{\lim }}\,\left( \frac{1}{{{\sin }^{2}}t{{\cos }^{2}}t}-\frac{1}{{{\sin }^{2}}t} \right)=\underset{t\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{2}}t}{{{\sin }^{2}}t{{\cos }^{2}}t}$

But ${{\sin }^{2}}t+{{\cos }^{2}}t=1\Leftrightarrow {{\sin }^{2}}t=1-{{\cos }^{2}}t$

So $3L=\underset{t\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}t}{{{\sin }^{2}}t{{\cos }^{2}}t}=\underset{t\to 0}{\mathop{\lim }}\,\frac{1}{{{\cos }^{2}}t}=1$ thus $L=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{\sin }^{2}}x}-\frac{1}{{{x}^{2}}}=\frac{1}{3}$