I suck at calculus...?

Question

..help me out!

Double check these answers please:

let f(x) = (3x^2 - 11x -20) / (x^2 +2x - 35)

(a)
lim->5 f(x) = 19/12

(b)
lim->4 f(x) = 16/11

(c)
lim->3 f(x) = 13/10

thankx

Answer 1

f(x) = [3x^2 - 11x - 20] / [x^2 + 2x - 35]

First, let's factor the top and bottom; this will be crucial to calculating the limit.

f(x) = [(3x + 4)(x - 5)] / [(x + 7)(x - 5)]

(a)

lim [(3x + 4)(x - 5)] / [(x + 7)(x - 5)]
x -> 5

Since x is close to but not equal to 5, we can cancel out the
(x - 5) on the top and bottom.

lim [(3x + 4)/(x + 7)]
x -> 5

Which we can now plug in safely.
(3(5) + 4) / (5 + 7) = (19)/(12) = 19/12

(b)

lim [(3x + 4)(x - 5)] / [(x + 7)(x - 5)]
x -> 4

lim [(3x + 4)] / [(x + 7)]
x -> 4

Since x = 4, then (3(4) + 4) / (3 + 7) = 16/10 = 8/5

(c)

lim [(3x + 4)(x - 5)] / [(x + 7)(x - 5)]
x -> 3

lim [(3x + 4)] / [(x + 7)]
x -> 3

(3(3) + 4) / (3 + 7) = 13/10

Answer 2

In calc, usually only the exponential terms are the ones you have to be concerned with when solving limits. For (a), think of it as 3(5^2)/(5^2). You would then have your answer.

If this is a simple plug and chug, your answers will work. If not maybe you should review your text and see what exactly they are looking for.