Question
Suppose f(x) = $ \left\{ {\begin{array}{*{20}{c}} {a + bx,} \\ {4,} \\ {b - ax,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 1} \\ {x = 1} \\ {x > 1} \end{array}$ and if $\mathop {\lim }\limits_{x \to 1}$ f(x) = f(1) what are possible values of a and b?

Answer

Here f(x) = $ \left\{ {\begin{array}{*{20}{c}} {a + bx,} \\ {4,} \\ {b - ax,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 1} \\ {x = 1} \\ {x > 1} \end{array}$
Also $\mathop {\lim }\limits_{x \to 1}$ f(x) = f(1)
$\Rightarrow \mathop {\lim }\limits_{x \to 1}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = f(1) = 4
$\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = 4 and $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = 4
Now $\mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ - }}$ (a + bx)
Put x = 1 - h as x $\to$ 1, h $\to$ 0
$\therefore \;\mathop {\lim }\limits_{h \to 0}$ [a + b(1 - h)] = $\mathop {\lim }\limits_{h \to 0}$ [a + b - bh] = a + b
Also $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ (b - ax)
Put x = 1 + h as x $\to$ 1, h $\to$ 0
$\therefore \;\mathop {\lim }\limits_{h \to 0}$ [a - b(1 + h)] = $\mathop {\lim }\limits_{h \to 0}$ [b - a - ah] = b - a
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

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