Question

1. ( f(x) = \frac{x^2 - 25}{x - 5} ); ( a = 5 )

Explanation:

Step1: Simplify the function

Notice that the numerator \(x^2 - 25\) is a difference of squares, which can be factored as \((x - 5)(x + 5)\). So we have:
\[f(x)=\frac{(x - 5)(x + 5)}{x - 5}\]
For \(x
eq5\), we can cancel out the common factor \((x - 5)\) in the numerator and the denominator, so \(f(x)=x + 5\) (when \(x
eq5\)).

Step2: Find the limit as \(x\) approaches \(a = 5\)

To find \(\lim_{x
ightarrow5}f(x)\), we can use the simplified function \(f(x)=x + 5\) (since the limit as \(x\) approaches 5 does not depend on the value of the function at \(x = 5\), only the values around \(x = 5\)). Substitute \(x = 5\) into \(x + 5\):
\[ \lim_{x
ightarrow5}f(x)=5 + 5=10\]

Answer:

The limit of \(f(x)\) as \(x\) approaches \(5\) is \(10\).