<p>how do you solve for limits when they approach zero?? i am confused, i get it when it is numbers and smts. infinity..
also, the l'hopital's rule, is only for Calc BC right?</p>
<p>Although L'Hopital's rule is part of the BC curriculum and not of the AB curriculum, the rule is so simple it can't hurt to know it, and it may prove useful to answer one or two questions on the exam. Just remember that if when taking a limit of one function over the other you get the indeterminate form, then the limit is equal to the derivative of the top one divided by the derivative of the bottom one.</p>
<p>Suppose, for example, that I wanted to find the following.</p>
<p>lim f(x) = (x/x)
as x -> 0</p>
<p>I could simply cancel and get 1, in which case I know the limit everywhere is 1, but lets avoid that for the purposes of illustrating L'Hopital's rule. Using said rule, you would take the derivative of x over the derivative of x. You end up with 1/1, so the limit is 1. This works since trying to take the limit at first produces the indeterminate form of 0/0.</p>
<p>As for taking limits as they approach 0, most of the time you just plug in the value into the equation. For example, consider. </p>
<p>lim f(x) = 3x+4+5x^2
x -> 0</p>
<p>The answer is 4. For continuous functions, the limit and the functional value is the same, so simply finding the functional value by evaluating the function when x = 0 suffices. Sometimes you get tougher questions, such as:</p>
<p>lim f(x) = (1/x)
x -> 0</p>
<p>You have a problem here. There is no functional value at 0 because the function 1/x isn't continuous at 0. In this particular example, the limit does not exist. If you take the right handed limit (as x approaches 0 from the right), you get positive infinity. If you take the left handed limit (as x approaches 0 from the left) you get negative infinity. Since the right handed limit is different from the left handed limit, the limit is said not to exist.</p>
<p>If you have any specific problems, please post them and I will try to help you with them ^.^</p>
<p>same way as you do for any other number...</p>
<p>example: lim(x>>>0) (x^3 - 9) / (x+3)
first expand the top, and cancel out the (x+3; so you are left with x-3 in the numerator)
now lim(x>>>0) x-3 = 0-3 = -3</p>
<p>IF YOU HAVE ANY SPECIFIC QUESTIONS, I WILL BE MORE THAN HAPPY TO HELP YOU OUT!!! :)</p>
<p>l'hopital is yes, only for BC, but it makes solving for limits VERY EASY if you know it; so I guess what I'm trying to say is, if you should know it regardless</p>
<p>EDIT: TOO LATE I GUESS :)</p>
<p>fadingaway, also note that in the above example you don't have to factor and cancel. Simply plugging in 0 gives (-9/3) = -3, which is exactly the same result (BTW, vader1990, I think you made a mistake... you mean x^2, right? Since you are factoring the difference of the squares, XD)</p>
<p>thanks for the examples...how do i go about solving this one w/o a calculator:
lim (x--> 0) sin (1/x)??
oh..and can someone explain the infinity type problems for example, lim (x---> infinity) 2x^2+1/(2-x)(2+x)?? thanks!!</p>
<p>lol...I was just trying to make some example up without thinking about it, guess I EPIC FAILED :) lol lol lol</p>
<p>OK, LET ME TRY TO SOMEWHAT REDEEM MYSELF...
sin(1/x) = limit doesn't exist
the left-hand limit x>>0- is -infinity
the right-hand limit x>>o+ is +infinity
since the limits don't match, the net limit is undefined!</p>
<p>well, there are two ways to do the infinity problems, you can either divide each term by the highest exponent, so do what I do and l'hopital it...
1) dividing each term by the largest exponent..but wait there is a quicker way to do this, since the power of x (2 in this case) is the same in the nemerator and denominator, you can jus tdivide the exponenets of the x^2, and that will be your answer--in this case 2 / -1 = -2
2) l'hopital, is just the derivative of the numerator, divided by the derivative of the denominator, yields 2 / -1 = -2</p>
<p>Actually, vader's right that the limit as x->0 of sin (1/x) doesn't exist, but not on the reason why. The biggest problem is that the right-hand limit is equivalent to lim y->infinity sin (y), which doesn't exist. (The left-hand limit is equivalent to lim y->-infinity sin (y), which also doesn't exist.)</p>
<p>okay thanks...i kinda get it..LoL..
i hav a question regarding ap calc in general..if i don't understand ANYTHING except for lik basic derivatives and integrals, if i spend 1-2 hrs a day starting this week, will i be able to get a 3+ on the Ap exam?...how should i go about studying..and which books should i get....??</p>
<p>bump....also, is it true that i only need 60% right to get a 5??</p>
<p>lol, "only" 60%? It seems to me you are severely underestimating the difficulty of the exam. Good luck getting that one; there is a reason they made that score a 5, you know. Besides, isn't it more like 70%?</p>
<p>Anyways, as for being able to get a 3 with reviewing and using basic derivatives and integrals... By basic, you mean the table of derivatives and integrals that appear on the front or back of the book that you can memorize? And not understanding anything else at all? I would guess that would get you a 2, but I don't know about a 3. Look at the sample questions offered by college board and tell me if you can do them.</p>
<p>we've done that in class, and i am able to do most of them...</p>
<p>What about the free responses? Look here and tell me how you do answering those. <a href="http://apcentral.collegeboard.com/apc/members/exam/exam_questions/1997.html%5B/url%5D">http://apcentral.collegeboard.com/apc/members/exam/exam_questions/1997.html</a></p>
<p>But, at the very least, doing good on most of the practice multiple choice questions is quite reassuring. In that case, a 3 might definitely be within your reach.</p>
<p>i can get some of them..i'm just not very confident all together, so i want to know what books can help me get a good review in well, less than a month..</p>
<p>Actually, l'Hopitals rule can be useful in AB too! our teacher taught it to us last week even though we are in AB.
Basically, if you do not know yet, if you get 0/0 or inf/inf when you plug in that number, you can take the derivative of the function and take the limit again. If it ends up indeterminate form again (o/o,etc..) then you do it over again.</p>