Question

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

Answer 1

Here f(x) = $\{\begin{array}{ccc}a+bx& x<1& \\ 4& x=1& \\ b-ax& x>1& \end{array}$

Also $\underset{x\to 1}{\text{lim}}$ f(x) = f(1)

$\Rightarrow \underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 1^{+}}{\text{lim}}$ f(x) = f(1) =4

$\Rightarrow \underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=4and\underset{x\to 1^{+}}{\text{lim}}$ f(x) = 4 ...(i)

Now $\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 1^{-}}{\text{lim}}$ (a+bx)

Put x = 1- h as $x\to 1,h\to 0$

$\therefore \underset{h\to 0}{\text{lim}}[a+b(1-h\left)\right]=\underset{h\to 0}{\text{lim}}$ [a+b-bh] = a+b ...(ii)

Also $\underset{x\to 1^{+}}{\text{lim}}f\left(x\right)=\underset{x\to 1^{+}}{\text{lim}}$ (b-ax)

Put x =1 + h ax $x\to 1,h\to 0$

$\therefore \underset{h\to 0}{\text{lim}}[b-a(1+h\left)\right]=\underset{h\to 0}{\text{lim}}$ [b-a-ah] = b-a ...(iii)

Putting values from (ii) and (iii) in(i)
$\therefore$ a+b= 4 and -a + b =4

Solving these equations, we have
a = 0 and b=4.