Factorize the denominator:
[tex]x^2-31x+240 = (x-16)(x-15)[/tex]
Then we find that ...
• When c = 15,
[tex]\displaystyle \lim_{x\to15}f(x) = \lim_{x\to15}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to15}\frac1{x-16} = \frac1{15-16} = \frac1{-1} = \boxed{-1}[/tex]
because the factors of x - 15 in the numerator and denominator cancel with each other. More precisely, we're talking about what happens to f(x) as x gets closer to 15, namely when x ≠ 15. Then we use the fact that y/y = 1 if y ≠ 0.
• When c = 16,
[tex]\displaystyle \lim_{x\to16}f(x) = \lim_{x\to16}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to16}\frac1{x-16} = \dfrac10[/tex]
which is undefined; so this limit does not exist.
• When c = 17,
[tex]\displaystyle \lim_{x\to17}f(x) = \lim_{x\to17}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to17}\frac1{x-16} = \frac1{17-16}=\frac11 =\boxed{1}[/tex]
because the function is continuous at x = 17.
