Answer to Question #308542 in Calculus for Amira

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

Solution (a)

$\begin{array}{l} f\left( x \right) = x + 2\\ \mathop {\lim }\limits_{x \to - 1} f\left( x \right) = \mathop {\lim }\limits_{x \to - 1} \left( {x + 2} \right) = - 1 + 2\\ f\left( { - 1} \right) = - 1 + 2\\ \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right) \end{array}\$

Solution (b)

$\begin{array}{l} f\left( x \right) = x - 2\\ \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( {x - 2} \right) = 0 - 2 = - 2\\ f\left( 0 \right) = 0 - 2 = - 2\\ \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right) \end{array}\$

Solution (c)

$\begin{array}{l} f\left( x \right) = & \left\{ \begin{array}{l} {x^2} - 1 & & x < - 1\\ {\left( {x - 1} \right)^2} -4& & x \ge - 1 & \end{array} \right. & c = - 1\\ \end{array}\$

$\begin{array}{l} \mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} \left( {{x^2} - 1} \right) = {\left( { - 1} \right)^2} - 1 = 0\\ \mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} {\left( {x - 1} \right)^2}-4 = {\left( { - 1 - 1} \right)^2} -4= 0 \end{array}\$

Since $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right)\$

Hence, the limit at $x=c=-1$ exists. 

Solution (d)

$\begin{array}{l} f\left( x \right) = & \left\{ \begin{array}{l} {x^3} - 1 & & x < 1\\ {x^2} + 4 & & x \ge 1 & \end{array} \right. & c = 1\\ \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^3} - 1} \right) = {\left( 1 \right)^3} - 1 = 0\\ \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} + 4} \right) = {\left( {{{\left( 1 \right)}^2} + 4} \right)^2} = 25 \end{array}\$

Since,

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\$

Hence, the limit at $x=c=1$ does not exist.