36 x − 2x^2 − 169 < h( x ) < x^2 − 18 x + 74

Question

36 x − 2x^2 − 169 < h( x ) < x^2 − 18 x + 74

We want to use the Squeeze Theorem to evaluate limx→9 h(x)

1.  First evaluate:  limx→9 36x−2x^2−169
2. Next evaluate:  limx→9. x^2−18x+ 74
3. Therefore, by the Squeeze Theorem, limx→9 h(x)=

Answer 1

Both functions equal -7 at 9

Now, the question is whether the function to the right is actually above the one to the left.

Subtract the first from the second, and we get

3x^2 -54x + 243 = 3(x^2 - 18x + 81) = 3(x-9)^2. The difference is greater than 0 everywhere except at 9, where it equals 0. This means that the statement is valid.

Thus, as h(x) is between two functions that have the same imit at 9, lim x -> 9 h(x) = -7

The statement is invalid at 9, of course.