Question

$$ f ( x ) = \frac { x - 4 } { x ^ { 2 } - 16 } $$ (a) explain why ( f ) has a removable discontinuity at ( x = 4 ). (select all that apply.) ( lim _ { x
ightarrow 4 } f ( x ) ) is finite. ( f ( 4 ) ) is undefined. ( f ( 4 ) ) and ( lim _ { x
ightarrow 4 } f ( x ) ) are finite, but are not equal. ( lim _ { x
ightarrow 4 } f ( x ) ) does not exists. none of the above nicely done. (b) redefine ( f ( 4 ) ) so that ( f ) is continuous at ( x = 4 ) (and thus the discontinuity is remove ( f ( 4 ) = 0.125 ) exactly!

Explanation:

Step1: Simplify the function

We know that \(x^{2}-16=(x + 4)(x - 4)\). So \(f(x)=\frac{x - 4}{x^{2}-16}=\frac{x - 4}{(x + 4)(x - 4)}\). For \(x
eq4\), we can cancel out the \(x - 4\) terms, and \(f(x)=\frac{1}{x + 4}\) when \(x
eq4\).

Step2: Calculate the limit

Now, we find \(\lim_{x
ightarrow4}f(x)\). Using the simplified form \(f(x)=\frac{1}{x + 4}\) (since the limit as \(x
ightarrow4\) does not consider the value of the function at \(x = 4\) itself), we substitute \(x = 4\) into \(\frac{1}{x+4}\). So \(\lim_{x
ightarrow4}f(x)=\frac{1}{4 + 4}=\frac{1}{8}=0.125\).

Step3: Check the function value at \(x = 4\)

The original function \(f(x)=\frac{x - 4}{x^{2}-16}\), when \(x = 4\), the denominator \(x^{2}-16=4^{2}-16=0\). So \(f(4)\) is undefined.

For a function to be continuous at \(x=a\), we need \(\lim_{x
ightarrow a}f(x)=f(a)\). Since \(\lim_{x
ightarrow4}f(x)\) is finite (\(0.125\)) and \(f(4)\) is undefined, by re - defining \(f(4)=\lim_{x
ightarrow4}f(x)\), we can make the function continuous at \(x = 4\).

Answer:

\(f(4)=0.125\)