Question

Please help. I don't understand what to input into the c...

Answer 1

Factorize the denominator:

$x^2-31x+240 = (x-16)(x-15)$

Then we find that ...

• When c = 15,

$\displaystyle \lim_{x\to15}f(x) = \lim_{x\to15}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to15}\frac1{x-16} = \frac1{15-16} = \frac1{-1} = \boxed{-1}$

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,

$\displaystyle \lim_{x\to16}f(x) = \lim_{x\to16}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to16}\frac1{x-16} = \dfrac10$

which is undefined; so this limit does not exist.

• When c = 17,

$\displaystyle \lim_{x\to17}f(x) = \lim_{x\to17}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to17}\frac1{x-16} = \frac1{17-16}=\frac11 =\boxed{1}$

because the function is continuous at x = 17.