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

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The given function is
$f(x)=\left\{\begin{array}{1 1}a+bx,&x<1\\4,&x=1 \\b-ax,&x>1\end{array}\right.$
$\lim\limits_{x \to 1^-}f(x) = \lim\limits_{x \to 1}(a+bx)=(a+b)$
$\lim\limits_{x \to 1^+}f(x) = \lim\limits_{x \to 1}(b-ax)=(b-a)$
$f(1)=4$
It is given that $\lim\limits_{x \to 1}f(x) = f(1)$
$\therefore \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x)= \lim\limits_{x \to 1} f(x) = f(1)$
$\Rightarrow a+b = 4 \: and\: b-a = 4$
On solving these two equations, we obtain $a = 0 \: and \: b = 4$
Thus, the respective possible values of $a$ and $b$ are $0$ and $4$
answered Apr 7, 2014

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