Question

To this function, does limit x approaching zero (without + or -) exist? Aside from that, is f(x) continuous at x=1?

Answer 1

Given:

$f(x)=\begin{cases}{-\frac{1}{x},x<0} \\ {3,0\leq x<1} \\ {\sqrt{x}}+2,x\ge1\end{cases}$

Required:

To find the limit exist if x approaching to 0.

Explanation:

As limit x approaching to 0,

$\lim_{x\rightarrow0}f(x)=3$

Therefore the limit exist.

$\begin{gathered} \lim_{x\rightarrow1^{-1}}f(x)=\lim_{x\rightarrow1^{-1}}3 \\ \\ =3 \end{gathered}$$\begin{gathered} \lim_{x\rightarrow1^+}f(x)=\lim_{x\rightarrow1^+}\sqrt{x}+2 \\ \\ =\sqrt{1}+2 \\ \\ =1+2 \\ \\ =3 \end{gathered}$