Answer to Question #305565 in Calculus for Luna Heart

Question

Determine whether if lim f(c) = f(c)

x + c

1. f(x) = x+2; c = -1

2. f(x) = x-2; c = 0

3. (at c = -1 )

f(x) = {x ² - 1 if x = < -1}

f(x) = { (x - 1) ² - 4 if x = ≥ - 1}

4. (at c = 1 )

f(x) = {x³ - 1 if x = < 1}

f(x) = { x² +4 if x = ≥ 1}

Answer 1

1, The function $f(x)=x+2$ is continuous on $\R$ as a polynomial.

Therefore

$$\lim\limits_{x\to c}f(x)=f(c), c\in \R$$

Then

$$\lim\limits_{x\to -1}f(x)=f(-1)=-1+2=1$$

2, The function $f(x)=x-2$ is continuous on $\R$ as a polynomial.

Therefore

$$\lim\limits_{x\to c}f(x)=f(c), c\in \R$$

Then

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

3.

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

$$\lim\limits_{x\to -1^+}f(x)=(-1-1)^2-4=0$$

Since

$$\lim\limits_{x\to -1^-}f(x)=0=\lim\limits_{x\to -1^+}f(x),$$

then $\lim\limits_{x\to -1}f(x)$ exists, and $\lim\limits_{x\to -1}f(x)=0.$

4.

$$\lim\limits_{x\to 1^-}f(x)=(1)^3-1=0$$

$$\lim\limits_{x\to 1^+}f(x)=(1)^2+4=5$$

Since

$$\lim\limits_{x\to 1^-}f(x)=0\not=5=\lim\limits_{x\to 1^+}f(x),$$

then $\lim\limits_{x\to 1}f(x)$ does not exist.