Question

If the function $f\left(x\right)$ defined as

$f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{(\sin x+\cos x)}^{\csc x},& -\frac{\pi }{2}<x<0\end{array}\\ \begin{array}{cc}a,& x=0\end{array}\\ \begin{array}{cc}\frac{e^{1/x}+e^{2/x}+e^{3/x}}{a{e}^{-2+1/x}+b{e}^{-1+3/x}},& 0<x<\frac{\pi }{2}\end{array}\end{array}$ is continuous at $x=0$, then

• A. $a=e,b=1$
• B. $a=1,b=e$
• C. $a=\frac{1}{e},b=1$
• D. None of these

Answer 1

The correct option is A $a=e,b=1$

We have,

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\left[\sin (-h)+\cos (-h)\right]}^{\csc h}$

$=\underset{h\to 0}{lim}{\left[(\cos h-\sin h)\right]}^{-\csc h}$

$=\underset{h\to 0}{lim}{(1+(\cos h-\sin h-1))}^{\frac{1}{\cos h-\sin h-1}.\frac{\cos h.\sin h-1}{-\sin h}}$

$={\left[\underset{y\to 0}{lim}{(1+y)}^{\frac{1}{y}}\right]}^{{lim}_{h\to 0}\frac{\cos h-\sin h-1}{-\sin h}}$

Now, $\underset{h\to 0}{lim}\frac{\cos h-\sin h-1}{-\sin h}\left(\frac{0}{0}\right)$

$=\underset{h\to 0}{lim}\frac{-\sin h-\cos h}{-\cos h}=\frac{0-1}{-1}=1$

Thus, $\underset{x\to 0^{-}}{lim}f\left(x\right)=e$

Now, we have,

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{e^{\frac{1}{h}}+e^{\frac{2}{h}}+e^{\frac{3}{h}}}{{ae}^{-2+\frac{1}{-h}}+{be}^{-1+\frac{3}{h}}}$

$=\underset{h\to 0}{lim}\frac{e^{\frac{-2}{h}}+e^{\frac{-1}{h}}+1}{\left({ae}^{-2}\right)e^{\frac{-2}{h}}+\left({be}^{-1}\right)}=\frac{0+0+1}{\left({ae}^{-2}\right)0+\left({be}^{-1}\right)}=\frac{e}{b}$

If $f$ is continuous at $x=0,$

then $e=a=\frac{e}{b}$ given $a=e$ and $b=1$