Question

Suppose f(x)={kx + 7 ; x>= 2 ,x^2+19 ; x < 2' , For what value of k is lim x→2 f(x) defined?

Answer 1

$$f(x) = \begin{cases} x^2+19 &x<2 \\ kx + 7 &x\geq 2 \end{cases}$$

$$\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}(x^2+19)=(2)^2+19=23$$

$$\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}(kx+7)=2k+7$$

If $\lim\limits_{x\to 2}f(x)$ is defined, then

$$\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2}f(x)$$

Hence

$$23=2k+7=>k=8$$

$\lim\limits_{x\to 2}f(x)$ is defined, if $k=8.$