# Calculus 1 (limits, VA and constants)

1. Jan 28, 2006

### gator

Im in calculus 1 and need some help

Question: Find the vertical asymptotes for…
F(x) = x^3 + 8 / x^2 –4
I got a va of –2

F(x) = 2x/ sin2x
I got a va of 1

Find the limit
Lim tan^2 2x/ x^2
x-0

I got 4 as my answer

Lim x^2 –4 / (x + 2 )^2
x- (-2)

I got –4/0 = no limit

Constant problem

F(x) = -x + 2 x<c
2x^2 –2x –4 x_>c

Not sure how to answer this.

Where can I find notes on this section online?
Thanks

2. Jan 28, 2006

### HallsofIvy

Did you notice this NOT the "homework" section?
PLEASE use parentheses! I assume you really meant
F(x)= (x^3+ 8)/(x^2- 4)= ((x+ 2)(x^2- 2x+ 4))/((x+2)(x-2))=
((x-2)(x-2))/(x-2)= x-2 as long as x is not 2 or -2. I don't see any vertical asymptote there.

F(1)= 2/sin 2. How is that "vertical"?

HOW did you get 4?

Once again, HOW did you get "-4/0"?

3. Jan 28, 2006

### VietDao29

As a matter of fact, $(x ^ 2 - 2x + 4) \neq (x - 2) ^ 2$
I think he probably meant:
$$\lim_{x \rightarrow -2} \frac{x ^ 2 - 4}{(x + 2) ^ 2} = \lim_{x \rightarrow -2} \frac{x - 2}{(x + 2)} \rightarrow \frac{-4}{0}$$
(@ gator, I wouldn't say -4 / 0 = no limit (as -4 / 0 is not really well-defined), I'd rather say, as x tends to -2, the numerator tends to -4, and the denominator tends to 0, hence there's no limit).
How much will a chance probably cost me? :tongue:

Last edited: Jan 28, 2006
4. Jan 29, 2006

### HallsofIvy

Well, dang me! It was probably that fourth whiskey and soda.