Calculus 1 (limits, VA and constants)

[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

 

[HallsofIvy]

Im in calculus 1 and need some help

Did you notice this NOT the "homework" section?

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

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

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

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

I got 4 as my answer

HOW did you get 4?

Lim x^2 –4 / (x + 2 )^2
x- (-2)

I got –4/0 = no limit

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

Constant problem

F(x) = -x + 2 x<c
2x^2 –2x –4 x_>c

Not sure how to answer this. [\quote]

Answer what? There is no question here!

If, buy chance, you mean "Find c so that F(x) is continuous for all x" then you should find the one-sided limits as x-> c. Since -x+ 2 and 2x^2- 2x- 4 are both polynomials (and so continuous) you can do that by just setting x= c in each. The function will be continuous at x= c if those are the same. (There are two answers.)

Where can I find notes on this section online?
Thanks

Since I really have no idea what "section" you are talking about, I don't know.

 

[VietDao29]

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.

As a matter of fact, (x ^ 2 - 2x + 4) \neq (x - 2) ^ 2

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

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).

If, buy chance, you mean "Find c so that F(x) is continuous for all x" then you should find the one-sided limits as x-> c. Since -x+ 2 and 2x^2- 2x- 4 are both polynomials (and so continuous) you can do that by just setting x= c in each. The function will be continuous at x= c if those are the same. (There are two answers.)

How much will a chance probably cost me?

 

[HallsofIvy]

As a matter of fact, (x ^ 2 - 2x + 4) \neq (x - 2) ^ 2

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