Determine the values of a, b and c for which the function :$f(x)= \left\{ \begin{array}{1 1} \Large\frac{\sin (a+1)x+\sin x}{x} & \quad ,\;x<0 \\ c & \quad ,\;x=0 \\ \Large\frac{\sqrt{x+bx^2}-\sqrt x}{b\sqrt{x^3}} & \quad ,\;x>0 \end{array} \right.$ may continuous at x = 0

Toolbox:
• For continuous at $x = \pi, LHL = RHL = f(\pi)$
$LHL \: \lim\limits_{x \to 0} \: \large\frac{sin(a+1)x+sinx}{x}=a+2$
$= \lim\limits_{x \to 0} \: \large\frac{sin(a+1)x*(a+1)}{x*(a+1)}+\lim\limits_{x \to 0} \: \large\frac{sinx}{x}$
$= a+1+1=a+2$
$RHL = \lim\limits_{x \to 0} \: \large\frac{\sqrt{x+bx^2}-\sqrt x}{b\sqrt{x^3}}=\frac{\large1}{\large2}$
$\lim\limits_{h \to 0} \: \large\frac{\sqrt x ( \sqrt{1+bx}-1)}{bx\sqrt x}$

Rationalising numerator
$\Rightarrow \lim\limits_{h \to 0} \: \large\frac{\not1+bx\: -\not1}{bx(\sqrt{1+bx}+1}$
$\Rightarrow \lim\limits_{h \to 0} \: \large\frac{1}{\sqrt{1+bx+1}}=\large\frac{1}{2}$
$f(0)=c$
f is continuous
$\Rightarrow a+2=\frac{1}{2}=c$

solving which $a=\large\frac{-3}{2}, c = \large\frac{1}{2}$
and b can be any real number other than 0

edited Mar 26, 2013