Question

Suppose $f\left(x\right)=\{\begin{array}{cc}a+bx,& x<1\\ 4,& x=1\\ b-ax,& x>1\end{array}$ and if $\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$ what are possible values of $a$ and $b?$

Answer 1

Step $1:$ Limit at $x=1$
Here, limit exists at $x\to 1$
i.e., $L.H.L.=R.H.L.=f\left(1\right)=4\cdots \left(i\right)$
$L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1-h)$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}a+b(1-h)$
$\Rightarrow L.H.L.=a+b(1-0)$
$\Rightarrow L.H.L.=a+b\cdots \left(ii\right)$

$R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1+h)$
$\Rightarrow R.H.L.=\underset{h\to 0}{lim}b-a(1+h)$
$\Rightarrow R.H.L.=b-a(1+0)$
$\Rightarrow R.H.L.=b-a\cdots \left(iii\right)$

Step 2: Solve equations for value of $a$ and $b$
Given, $\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$
$\Rightarrow \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)$
$\Rightarrow a+b=b-a=4$

From $\left(ii\right)$ and $\left(iii\right)$
$\therefore a+b=4\cdots \left(iv\right)$
$b-a=4\cdots \left(v\right)$
Adding equation $\left(iv\right)$ and $\left(v\right)$
$\Rightarrow a+b+b-a=4+4$
$\Rightarrow 2b=8\Rightarrow b=4$
Also, $a+b=4$
$\Rightarrow a+4=4$
$\Rightarrow a=0$