Question

Find k so that $\underset{x\to 2}{\lim }f\left(x\right)$ may exist, where $f\left(x\right)=\left\{\begin{array}{ll}2x+3,& x\le 2\\ x+k,& x>2\end{array}\right..$

Answer 1

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}2x+3,& x\le 2\\ x+k,& x>2\end{array}\right. \\ \mathrm{Left}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 2^{-}}{\lim }f\left(x\right) \\ =\underset{x\to 2^{-}}{\lim }\left(2x+3\right) \\ \mathrm{Let}x=2-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[2\left(2-h\right)+3\right] \\ =2\left(2-0\right)+3 \\ =7 \\ \mathrm{Right}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 2^{+}}{\lim }f\left(x\right) \\ =\underset{x\to 2^{+}}{\lim }\left(x+k\right) \\ \mathrm{Let}x=2+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left(2+h+k\right) \\ =2+k \end{gathered}$$

Now, $\underset{x\to 2}{\lim }f\left(x\right)$ exists if the left hand limit is equal to the right hand limit.

⇒7 = 2 + k
k = 5