Question

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sin (a+1)x}{x}& ,ifx<0\\ c& ,ifx=0\\ \frac{\sqrt{x+b{x}^2}-\sqrt{x}}{b{x}^{3/2}}& ,ifx>0\end{array}\right\}$ is continuous at $x=0$ find $a,b$ and $c$. then a+c =

Answer 1

$f\left(x\right)=$ { $\frac{sin(a+1)x}{x},x<0$

{ $C,x=0$

{ $\frac{\sqrt{x+{bx}^2}-\sqrt{x}}{b{x}^{3/2}},x>0$

Given $f\left(x\right)$ is continuous at $x=0$

The left hand limit at $x=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(o-h)=\underset{h\to 0}{lim}\frac{\sin (a+1)h}{h}=\underset{h\to 0}{lim}\frac{\sin (a+1)h}{h}$

$=\underset{h\to 0}{lim}$ $\frac{\sin (a+1)h}{(a+1)h}(a+1)$

$=(a+1)\underset{h\to 0}{lim}$ $\frac{\sin (a+1)h}{(a+1)h}$

$=(a+1)$ $(\because \underset{x\to 0}{lim}\frac{sinx}{x}=1)$

The right hand limit $a+x=0$

$\underset{x\to 0^{-}}{lim}$ $f\left(x\right)=\underset{h\to 0}{lim}$ $f(o+h)=\underset{h\to 0}{lim}$ $\frac{\sqrt{h+{bh}^2}-\sqrt{h}}{b{h}^{3/2}}$

$=\underset{h\to 0}{lim}$ $\frac{\sqrt{h}\sqrt{1+bh}-\sqrt{h}}{bh.\sqrt{h}}$

$=\underset{h\to 0}{lim}$ $\frac{\sqrt{1+bh}-1}{bh}$

$=\underset{h\to 0}{lim}$ $\frac{\frac{1}{2}{(1+bh)}^{-1/2}\left(b\right)}{b}$ (Using $L^{\prime }$ Hospital's rule)

$=\underset{h\to 0}{lim}$ $\frac{1}{2\sqrt{1+bh}}=\frac{1}{2}$

Since the function is continuous at $x=0$,

$\therefore$ the left hand limit $=$ right hand limit $=$ functional value

$\Rightarrow a+1=1/2$

$\Rightarrow a=-1/2$

also $c=1/2$

and $bϵR$

$\therefore$ $a=-1/2$ $c=1/2$ $bϵR$.