Question:
Suppose $f(x)=\left\{\begin{array}{ll}a+b x, & x<1 \\ 4, & x=1 \\ b-a x & x>1\end{array}\right.$ and if $\lim _{x \rightarrow 1} f(x)=f(1)$ what are possible values of $a$ and $b ?$
Solution:
The given function is
$f(x)= \begin{cases}a+b x, & x<1 \\ 4, & x=1 \\ b-a x & x>1\end{cases}$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(a+b x)=a+b$
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(b-a x)=b-a$
$f(1)=4$
It is given that $\lim _{x \rightarrow 1} f(x)=f(1)$.
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{\left.x \rightarrow\right|^{+}} f(x)=\lim _{x \rightarrow 1} f(x)=f(1)$
$\Rightarrow a+b=4$ and $b-a=4$
On solving these two equations, we obtain $a=0$ and $b=4$.
Thus, the respective possible values of $a$ and $b$ are 0 and 4