Question

lf $f\left(x\right)=\frac{\tan x}{\sqrt{1+{\tan }^{2}x}},\underset{x\to (\pi /2{)}^{-}}{Lim}f\left(x\right)=a$ and $\underset{x\to (\pi /2{)}^{+}}{Lim}f\left(x\right)=b$ then

• A. $a=b$
• B. $a=1+b$
• C. $a+b=0$
• D. $a+b=2$

Answer 1

Answer: (C)

$a+b=0$

Given $f\left(x\right)=\frac{\tan x}{\sqrt{1+{\tan }^{2}x}}$

$\underset{x\to (\pi /2{)}^{-}}{Lim}f\left(x\right)=a$

$\underset{x\to (\pi /2{)}^{+}}{Lim}f\left(x\right)=b$

$f\left(x\right)=\frac{\tan x}{\sqrt{{\sec }^{2}x}}$ $[\because {\sec }^{2}x=1+{\tan }^{2}x]$

$=\frac{\tan x}{|\sec x|}$

$a=\underset{x\to (\pi /2{)}^{-}}{Lim}f\left(x\right)$

$=\underset{x\to (\pi /2{)}^{-}}{Lim}\frac{\tan x}{|\sec x|}$

here $x\to (\pi /2{)}^{-}$ i.e., $x<\frac{\pi }{2}$ which lies in $1^{st}$ Quadrant

So, $|\sec x|=\sec x$

$\Rightarrow a=\underset{x\to (\pi /2{)}^{-}}{Lim}\frac{\tan x}{\sec x}$

$=\underset{x\to (\pi /2{)}^{-}}{Lim}\frac{\sin x}{\cos x.\sec x}$

$=\underset{x\to (\pi /2{)}^{-}}{Lim}\sin x$

$=\sin \frac{\pi }{2}$

$=1$

Similarly

$b=\underset{x\to {\left(\frac{\pi }{2}\right)}^{+}}{Lim}\frac{\tan x}{|\sec x|}$

here

$x\to {\left(\frac{\pi }{2}\right)}^{+}$ i.e., $x>\frac{\pi }{2}$ which lies in $2^{nd}$ Quadrant $|\sec x|=-\sec x$

$\Rightarrow b=\underset{x\to {\left(\frac{\pi }{2}\right)}^{+}}{Lim}\frac{\tan x}{-\sec x}$

$=\underset{x\to {\left(\frac{\pi }{2}\right)}^{+}}{Lim}\frac{\sin x}{\cos x\times (-\sec x)}$

$=\underset{x\to {\left(\frac{\pi }{2}\right)}^{+}}{Lim}-\sin x$

$=-\sin \pi /2$

$=-1$

$a+b=1+(-1)=0$

$\therefore a+b=0$