Question

If $\begin{align}
  & f(x)=\dfrac{\sin (a+1)x+2\sin x}{x},x<0 \\
 & =2,\text{ }x=0 \\
 & =\dfrac{\sqrt{1+bx}-1}{x},\text{ }x>0 \\
\end{align}$
Is continuous at $x=0,$then find the values of $a\text{ and }b$.

Answer 1

Hint:In the given question the given function is defined in three part on a number line that is $x=0,x>0\text{ and }x<0$, and it is given that the function is continuous at $x=0$. As the function is continuous at $x=0$, so left hand limit, right hand limit and value of function at $x=0$, all are equal.
So, find $f(0-),f(0+)\text{ and f(0)}$. Once we find these values, we have to form equation and solve it in order to get the value of $a\text{ and }b$.

Complete step by step answer:
We have given
$\begin{align}
  & f(x)=\dfrac{\sin (a+1)x+2\sin x}{x},x<0 \\
 & =2,\text{ }x=0 \\
 & =\dfrac{\sqrt{1+bx}-1}{x},\text{ }x>0 \\
\end{align}$
Now we have to find left-hand limit
\[\begin{align}
  & \underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{\sin (a+1)x+2\sin x}{x} \\
 & \underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{\sin (a+1)x}{x}+\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{2\sin x}{x} \\
\end{align}\]
As we know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$
We can write the above limit as
\[\begin{align}
  & \underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{(a+1)\sin (a+1)x}{(a+1)x}+2\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{\sin x}{x} \\
 & \underset{x\to 0-}{\mathop{\lim }}\,f(x)=(a+1)\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{\sin (a+1)x}{(a+1)x}+2\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{\sin x}{x} \\
 & \underset{x\to 0-}{\mathop{\lim }}\,f(x)=(a+1)+2 \\
 & \underset{x\to 0-}{\mathop{\lim }}\,f(x)=a+3-------(1) \\
\end{align}\]
Hence, we get the value of left- hand limit
Now we have from question
$f(0)=2------(2)$
Now we have to find right-hand limit
$\underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{x\to 0+}{\mathop{\lim }}\,\dfrac{\sqrt{1+bx}-1}{x}$
Now we can rationalize the limit and can write
\[\begin{align}
  & \underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{x\to 0+}{\mathop{\lim }}\,\left( \dfrac{\sqrt{1+bx}-1}{x} \right)\left( \dfrac{\sqrt{1+bx}+1}{\sqrt{1+bx}+1} \right) \\
 & \underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{x\to 0+}{\mathop{\lim }}\,\left( \dfrac{1+bx-1}{x\left( \sqrt{1+bx}+1 \right)} \right) \\
 & \underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{x\to 0+}{\mathop{\lim }}\,\left( \dfrac{bx}{x\left( \sqrt{1+bx}+1 \right)} \right) \\
\end{align}\]
Now we have to cancel the terms $x$ and put the value $x=0$to get the right-hand limit, hence we can write
\[\begin{align}
  & \underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{x\to 0+}{\mathop{\lim }}\,\left( \dfrac{b}{\left( \sqrt{1+bx}+1 \right)} \right) \\
 & \underset{x\to 0+}{\mathop{\lim }}\,f(x)=\dfrac{b}{2}---------(3) \\
\end{align}\]
As function is continuous as $x=0$we can write from $(1),(2)\text{ and (3)}$
$a+3=2=\dfrac{b}{2}$
Hence, we get the value
$\begin{align}
  & a=3 \\
 & b=4 \\
\end{align}$

Note:
The geometrical meaning of continuity of a function at any point is that, if we draw the graph at the point of continuity, it must not be break. It should be smoothly drawn. If the graph broken it means it is discontinuous at that point.
A function is discontinuous at $x=a$ in each of the following case
Case 1. When $f(a)$is not defined
Case 2. When $\underset{x\to a}{\mathop{\lim }}\,f(x)$does not exist
Case 3. When $\underset{x\to a}{\mathop{\lim }}\,f(x)\ne f(a)$
The third case is removable discontinuity.