Proving continuity

Question

F(x)={x^2-3x-10/x-5; x doesn't equal 5

        {7; x=5

a) limx->5-

b) limx->5+

c) is the function continuous at x=5

 

i got 0, 7, and no. I'm both scared to answer the question and unsure if this is right.

Answer 1

f(x ) = (x² - 3x - 10)/(x - 5) if x ≠ 5 and f(5) = 7 

 

If x ≠ 5, then (x² - 3x - 10)/(x - 5) = (x - 5)(x + 2)/ (x - 5) = x + 2

 

lim f(x) as x → 5^- = 5 + 2 = 7

lim f(x) as x → 5⁺ = 5 + 2 = 7

 

Since the one-sided limits are equal, lim f(x) as x → 5 = 7 = f(5).

 

Since f(5) is defined, lim as x→5 exists and lim as x→5 = f(5), f(x) is continuous at x = 5.