Limit Exercises without using L'hopital Rule

Exercise:

Show that , $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\cos \sqrt{x}}-1}{x}=\frac{-1}{4}$

Solution: we have $\frac{\sqrt{\cos \sqrt{x}}-1}{x}=\frac{\sqrt{\cos \sqrt{x}}-1}{x}\times \frac{\sqrt{\cos \sqrt{x}}+1}{\sqrt{\cos \sqrt{x}}+1}=\frac{\cos \sqrt{x}-1}{x\left( \sqrt{\cos \sqrt{x}}+1 \right)}$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\cos \sqrt{x}}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \sqrt{x}-1}{x\left( \sqrt{\cos \sqrt{x}}+1 \right)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \sqrt{x}-1}{x}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{\cos \sqrt{x}}+1}$

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \sqrt{x}-1}{x}\times \frac{1}{2}=\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \sqrt{x}-1}{x}$

We know that ${{\sin }^{2}}\theta =\frac{1}{2}\left( 1-\cos 2\theta  \right)\Leftrightarrow 2{{\sin }^{2}}\left( \frac{\theta }{2} \right)=1-\cos \theta $

put $\theta =\sqrt{x}$

So $\cos \sqrt{x}-1=-\left( 1-\cos \sqrt{x} \right)=-2{{\sin }^{2}}\left( \frac{\sqrt{x}}{2} \right)$

Thus \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \sqrt{x}-1}{x}=-\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( \frac{\sqrt{x}}{2} \right)}{x}=-\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( \frac{\sqrt{x}}{2} \right)}{{{\left( \sqrt{x} \right)}^{2}}}\)

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

Thus $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\cos \sqrt{x}}-1}{x}=\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \sqrt{x}-1}{x}=\frac{1}{2}\times \frac{-1}{2}=\frac{-1}{4}$