Answer:
A. Yes, the point is defined, limit f(x) exists, and limit f(x) 0.
Step-by-step explanation:
Let be [tex]f(x) = \frac{x^{2}-25}{x-7}[/tex], a rational function is continuous at a value of [tex]x[/tex] if and only if denominator is different from zero. In particullar, [tex]x-7 \neq 0[/tex].
Then, we find that function is continuous for every value of [tex]x[/tex] except [tex]x = 7[/tex]. Thus, the function is continuous at [tex]x = 0[/tex], which is evaluated below:
[tex]f(x) = \frac{(x+5)\cdot (x-5)}{x-7}[/tex]
[tex]f(0) = \frac{(0+5)\cdot (0-5)}{0-7}[/tex]
[tex]f(0) = \frac{(5)\cdot (-5)}{-7}[/tex]
[tex]f(0) = \frac{-25}{-7}[/tex]
[tex]f(0) = \frac{25}{7}[/tex]
Thus, correct answer is A.
