Limits In AP Calculus?

Question

How do you solve them? If the limit approches some number do you just plug that number into the equation and get it that way? If the limit you're trying to find goes to infinity then you find the horizontal asymtope right?

Answer 1

I've not taken the AP Exam, but that's really the only way to solve limits.

For limits approaching finite values, try these in order - if one doesn't work, try the next one:
1) Plugging in
2) Factor the function
3) Rationalize the function by multiplying by a conjugate of the numerator/denominator

Limits to infinity themselves tell you about horizontal asymptotes...I don't know how much you've covered of limits to infinity, but typically, you just start by plugging in the infinities. If you know the relative shapes of the graphs of the functions in question, that helps too. In the cases of indeterminate limits, you just use L'Hopital's Rule...

But yes, you're pretty much on the right track with that

Answer 2

I am not sure you are in AP CALCULUS AB or BC. Even though the l'Hopital's rule is not required for AP CALCULUS AB, I strongly suggest you learn it. Pay attention to the requirement for using l'Hopital's rule: 0/0 and infinity/infinity.

If the limit of the variable x goes to infinity, and the function f(x) goes to a constant, then you have a horizontal asymptote.

Answer 3

Sometimes you can do it that way, but alot of times you have to manipulate the equation some because if you plug in the number it ends up being divided by 0 or some such. I can't remember alot of it or i would help out more, sorry.

Answer 4

All of those are easily an identical. basically use algebra and factoring. e) actuality (sqrt(x) - 3) * (sqrt(x) + 3) = (sqrt(x))^2 - 3^2 = x - 9 shrink (9-x) / 2sqrt(x) -6) = (9-x) / (2)(sqrt(x) - 3) x--9 From the actuality we now get (3-sqrt(x)) * (3 + sqrt(x)) / -[2 * (3 - sqrt(x))] After the cancellation we've [3 + sqrt(x)] / 2 shrink [3 + sqrt(x)] / 2 = (3 + 3) / 2 = 6/2 = 3 x-->9 all something use algebra basically like this. for the reason which you asked 4 questions in a single, i'm going to flow away something to you. they're all easily an identical.

Answer 5

Limit= the number the equation gets closest to at point c

For example if f(x)=x
and c=1
Then the limit is 1 because it get closer to 1 when x gets closer to c

Answer 6

yu can not always plug in the no. ..yu have to check if the final result is defined or not...
and to the second one it is not always horizontal....depends on the function