Explain why the function f(x)={x^2-9}/{x-3} is discontinuous at x=3.
because it is undefined at x = 3.
The function f(x) = (x^2-9)/(x-3) is discontinuous at x = 3 due to the presence of a removable discontinuity. This means that as x approaches 3, the function is undefined or has a different value than what would be obtained by direct substitution.
To understand why the function is discontinuous at x = 3, we need to evaluate the limit of f(x) as x approaches 3 from both the left and the right-hand sides.
Let's first consider the limit as x approaches 3 from the left (x < 3). In this case, we substitute x = 3 - h, where h approaches 0. So, we have:
lim (x->3-) (x^2-9)/(x-3) = lim (h->0) ((3-h)^2-9)/(3-h-3)
Simplifying, we get:
lim (h->0) ((9 - 6h + h^2) - 9)/(-h) = lim (h->0) (-6h + h^2)/(-h)
Dividing numerator and denominator by -h, we have:
lim (h->0) (6 - h) = 6
Now let's consider the limit as x approaches 3 from the right (x > 3). In this case, we substitute x = 3 + h, where h approaches 0. So, we have:
lim (x->3+) (x^2-9)/(x-3) = lim (h->0) ((3+h)^2-9)/(3+h-3)
Simplifying, we get:
lim (h->0) ((9 + 6h + h^2) - 9)/(h) = lim (h->0) (6h + h^2)/(h)
Dividing numerator and denominator by h, we have:
lim (h->0) (6 + h) = 6
As we can see in both cases, the limits of f(x) as x approaches 3 from the left and from the right are both equal to 6. However, the function value at x = 3, obtained by direct substitution, is undefined because it would involve dividing by zero.
This discontinuity at x = 3 is known as a removable discontinuity because it can be fixed by redefining the function at that point. We can define f(x) = 6 for x = 3 and thus remove the discontinuity.