Question

Tell whether if the following piecewise function is a continuous at a given point or not. (SHOW THE SOLUTION).

1. at x = 4

x + 1 if x < 4

(x - 4)² + 3 if x ≥ 4

2. at x = 0

1/x if x ≤ -1

x² - 2 if x > -1

3. at x = 5

x² - 3 if x ≥ 5

4x + 2 if x < 5

Answer 1

1.

$$\lim\limits_{x\to4^-}f(x)=\lim\limits_{x\to4^-}(x+1)=4+1=5$$

$$\lim\limits_{x\to4^+}f(x)=\lim\limits_{x\to4^+}((x-4)^2+3)=(4-4)^2+3=3$$

$$\lim\limits_{x\to4^-}f(x)=5\not=3=\lim\limits_{x\to4^+}f(x)$$

$$\lim\limits_{x\to4}f(x)=\text{does not exist}$$

The function $f(x)$ is not continuous at $x=4.$

The function $f(x)$ has a jump discontinuity at $x=4.$

2.

$$\lim\limits_{x\to0}f(x)=\lim\limits_{x\to0}(x^2-1)=(0)^2-1=-1=f(0)$$

The function $f(x)$ is continuous at $x=0.$

3.

$$\lim\limits_{x\to5^-}f(x)=\lim\limits_{x\to5^-}(4x+2)=4(5)+12=22$$

$$\lim\limits_{x\to5^+}f(x)=\lim\limits_{x\to5^+}(x^2-3)=(5)^2-3=22$$

$$\lim\limits_{x\to5^-}f(x)=22=\lim\limits_{x\to5^+}f(x)=>\lim\limits_{x\to5}f(x)=22$$

$$f(5)=(5)^2-3=22=\lim\limits_{x\to5}f(x)$$

The function $f(x)$ is continuous at $x=5.$