\(f(x)=\frac{sin(a+1)x+sin x}{x},x<0.\)
\(f(x)=c,x=0\)
\(f(x)=\frac{\sqrt{x+bx^2}-\sqrt x}{bx^{3/2}}\;x>0\)
\(f(x) \) is acts at \(x=0.\)
LHL at \(x=0\) is \(f(0^-)=\) \([ \lim_{x\to0} \frac{sin(a+1)(0-h)+sin(0-h)}{0-h}\)
\(\lim_{x\to0} \frac{-sin(a+1)h-sin h}{h}\)
\(=\lim_{x\to0}(a+1)\frac{sin(a+1)h}{(a+1)}+\lim_{x\to0}\frac{sin h}{h}\)
\(=a+1+1(\because \lim_{x\to0} \frac{sin ax}{ax}=1)\)
\(=a+2\)
RHL of \(f(x) \) at \(x=0\) is
\(f(0^+)=\lim_{x\to0} \frac{\sqrt{h+bh^2}-\sqrt {h}}{bh^{3/2}}\) ( Replacing \(x\) by \(0+h=h\) )
\(=\lim_{h\to0}\frac{h^{1/2}(\sqrt{1+bh}-1)}{bh^{3/2}}\)
\(=\lim_{h\to0}\frac{\sqrt{1+bh}-1}{bh}\) ( \(\frac{0}{0}\) case )
\(=\lim_{h\to0}\frac{b}{2\sqrt{1+bh}}\times\frac{1}{b}\) ( By DLH Rule)
\(=\lim_{h\to0}\frac{1}{2\sqrt{l+b\times 0}}=\frac{1}{2}\)
\(\because f(x)\) are continuous at \(x= 0\)
\(\because f(0^-)=f(0^+)=f(0)\)
\(\because f(0)=C=\frac{1}{2}.\)
\(=a+2=\frac{1}{2}\)
\(\Rightarrow a=\frac{1}{2}-2\)
\(=\frac{-3}{2}.\)
Thus \(a=\frac{-3}{2},b=1\) and \(C=\frac{1}{2}.\)
