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)(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)(x-15)/((x-16)(x-15)) = \lim_(x\to16)\frac1{x-16} = \frac10`

which is undefined; so this limit does not exist.

• When c = 17,

`\displaystyle \lim_(x\to17)f(x) = \lim_(x\to17)(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.