Question

for the function ( f(x) ) shown below, determine ( limlimits_{x \to 3} f(x) ).

f(x) = \begin{cases} -9 + 4x & \text{for } x < 3 \\ -6 + x^2 & \text{for } x > 3 end{cases}

Explanation:

Step1: Find left - hand limit (x→3⁻)

To find the left - hand limit as \(x\) approaches 3, we use the part of the function for \(x < 3\), which is \(f(x)=- 9 + 4x\).
We substitute \(x = 3\) into this expression: \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{-}}(-9 + 4x)\)
Using the limit rule \(\lim_{x
ightarrow a}(c+mx)=c + ma\) (where \(c=-9\), \(m = 4\), \(a = 3\)), we get \(-9+4\times3=-9 + 12 = 3\).

Step2: Find right - hand limit (x→3⁺)

To find the right - hand limit as \(x\) approaches 3, we use the part of the function for \(x>3\), which is \(f(x)=-6 + x^{2}\).
We substitute \(x = 3\) into this expression: \(\lim_{x
ightarrow3^{+}}f(x)=\lim_{x
ightarrow3^{+}}(-6 + x^{2})\)
Using the limit rule \(\lim_{x
ightarrow a}(c + x^{2})=c+a^{2}\) (where \(c=-6\), \(a = 3\)), we get \(-6+3^{2}=-6 + 9=3\).

Step3: Determine the limit as x→3

Since the left - hand limit \(\lim_{x
ightarrow3^{-}}f(x) = 3\) and the right - hand limit \(\lim_{x
ightarrow3^{+}}f(x)=3\), and for a limit \(\lim_{x
ightarrow a}f(x)\) to exist, \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)\).
So, \(\lim_{x
ightarrow3}f(x)=3\).

Answer:

\(3\)