Question

Solve $\underset{x\to \pi }{lim}\frac{1+\cos x}{{\tan }^{2}x}$

• A. $\frac{1}{3}$
• B. $\frac{5}{2}$
• C. $\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

$\underset{x\to \pi }{lim}\frac{1+\cos x}{{\tan }^{2}x}$

Put $x=\pi +h$

$\underset{h\to 0}{lim}\frac{1+\cos (\pi +h)}{{\tan }^2(\pi +h)}=\underset{h\to 0}{lim}\frac{1-\cos h}{{\tan }^{2}h}$

We can say that $\cos h=1-2{\sin }^2\frac{h}{2}$

$1-\cos h=2{\sin }^2\frac{h}{2}$

$=\underset{h\to 0}{lim}\frac{2{\sin }^2\frac{h}{2}}{\frac{{\sin }^{2}h}{{\cos }^{2}h}}=2{\cos }^{2}h\underset{h\to 0}{lim}{\left(\frac{\sin \frac{h}{2}}{\sin h}\right)}^2$

$=2\underset{h\to 0}{lim}{(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{h}{\sin h}\times \frac{1}{2})}^2$

We know that $\underset{h\to 0}{lim}\frac{\sin h}{h}=1$

$=2\times {\left(\frac{1}{2}\right)}^2=2\times \frac{1}{4}=\frac{1}{2}$.