What does this function show at x = 5? f(x)=x^2−25/x−5

A) removable discontinuity
B) jump discontinuity
C) infinite discontinuity
D) continuity
E) none of the above

at x=5, f(x) = 0/0, or undefined.

At any other value, though,
f(x) = (x-5)(x+5)/(x-5) = x+5

So, f(x) is undefined only at x=5. By defining f(5) = 10, the discontinuity is removed. f(x) = x+5 for all x.

So would it be none of the above?

Ummm. Do you not think it would be A?

I explained how the discontinuity could be removed. Hence, it is removable.

To determine what the function shows at x = 5, we need to evaluate the function at that point.

The given function is f(x) = (x^2 - 25) / (x - 5).

To evaluate the function at x = 5, substitute 5 into the function:

f(5) = (5^2 - 25)/(5 - 5)

Evaluating the numerator:

= (25 - 25) / (5 - 5)
= 0 / 0

At this point, we cannot divide by zero, which results in an indeterminate form.

By observing the function, we can see that (x - 5) is a factor in both the numerator and denominator. Therefore, we can simplify the function by factoring:

f(x) = (x + 5)(x - 5)/(x - 5)

Now, the (x - 5) terms cancel out:

f(x) = (x + 5), where x ≠ 5

We've simplified the function to f(x) = x + 5, with the condition that x ≠ 5.

This means that at x = 5, the function is undefined as a result of the division by zero.

Therefore, the correct answer is: A) removable discontinuity