<p>For the people doing limits right now: just use L’Hopital’s rule, it’ll make your life a lot easier (or if your teacher will mark your answer wrong because of the approach, use it to check your answer.</p>
<p>It’s not to bad right now. What’s the L’Hopital’s rule?</p>
<p>isn’t L’Hospital’s rule BC calc?</p>
<p>After Tuesday we’ll finally be done with trig (thank goodness). I feel like I’m gonna ace this test. Been studying for the past hour and a half and I’m feeling good about it. Still got another day to ask him questions too. Can’t wait to get this trig stuff over with. So ready for the real calculus!
Are you guys’s classes getting super hard now? The BC class at my school is already complaining about how hard their class is</p>
<p>Yes it is, but who cares. It’s easy to learn and it works. (It probably won’t be useful on the AP exam as the exam development committee tends to avoid question that would give some students an unfair advantage, but it’s useful in class for homework/tests).</p>
<p>Oh ok, I think I am going to ttry to learn L’Hospital’s rule. Also to jspeed12, in my school, we are done with limits and we are learning derviatives. So far it seems easy. But my teacher is notorious for teaching easy lessons and giving super hard tests.</p>
<p>We just started derviatives at my school to. I’m not finding it complicated yet. This is the first year my teacher teaching AP so our notes are the what he gave his other class, then we get harder examples. The other Calc classes in my school use the same book as the AP class so it’s the same material with not as much getting done.</p>
<p>Nice nice I hated limits and derivatives last year when we did them so its gonna be even more fun doing them this year. I think we finally start limits tomorrow.
but yea big trig test tomorrow. Went to him for some extra help and I wouldn’t be surprised at all if I get 100 on this thing. He gave me one of the hardest problems he could think of and it took some time, but eventually I got it right and he said there aren’t gonna be any nearly that hard on the test so im pretty much ready.
Are you guys majoring in something math related as well? I’m doing chemical engineering</p>
<p>Junior, Calculus AB</p>
<p>I have a comprehensive derivatives test tomorrow. Should be easy. My tests are easy but the quizes are hard.</p>
<p>I am not worried about this AP though because last year, we had 27 take the test, 15 got 5s and the rest got 4s; the teacher is really good with practices and everything so got this covered.</p>
<p>In terms of major, I currently want to do computer engineering, but I really want to become an actuary which I have heard is super-math-intensive.</p>
<p>Only got a 95 on my test because of really careless mistakes. Kinda makes me a little angry. But we’re doing limits nose and we’ve been doing some with trig functions and now im sorta realizing why we spent so much time on trig. Also learned about the squeeze theorem.
Starting to really enjoy this class</p>
<p>Junior, Calculus AB</p>
<p>Got a 89% on my limits test, she put in a few curve balls so i was confused about them :S</p>
<p>started derivatives yesterday</p>
<p>I got As on my test but my teacher is awesome.</p>
<p>Sent from my HTC Glacier using CC App</p>
<p>The hardest part about derivatives is the algebra involved. My teacher constantly saying how were done with the calc in the first thirty seconds and are left with nasty trig and algebra problems.</p>
<p>well derivatives are actually easy if u know the shortcuts</p>
<p>My school only had 4 4’s on this test out of more than a hundred kids (the rest 5’s) although the downside to that is it just shows how difficult the class is.</p>
<p>i just had the easiest AP calc test ever on limits, but I made a stupid mistake. I thought of pi/2- as negative pi/2 instead of approaching pi/2 from the negative side.</p>
<p>I know the derivatives short cuts but our teacher won’t let us use them since he wants as to practice doing them the long way but it’s an easy way to check my work. CSMathAsa I hate when I make mistakes like that, it’s like I know that the anwser should be have been this but I have no clue why I put that last year on a test I said that 4*0 was 4 instead of 0. I was so mad when I got that test back</p>
<p>We have a test on limits on Tuesday and today we talked about continuity and removable discontinuity and literally everybody was like “lolwut…?”</p>
<p>Sophomore</p>
<p>Up to Implicit Differentiation (i think that’s what it’s called)
Inverse Trig Derives and Inverse Derivs</p>
<p>Supposedly the hardest section of the entire year, but was really easy for me. I just have a serious problem with trig, which could be my downfall if I don’t understand it soon. Thank goodness for Khan Academy. Thanks for posting the link!</p>
<p>We have a test on all the derivative rules sometime next week. Nervous because there will be a lot of trig. Hoping for an A this time</p>
<p>Since my class is done with limits here is a stduy guide for the limits unit(this is for people who haven’t taken a limits test yet or they will forget limits at the end of the year)
Limits and Continuity:
These can be solved in 3 ways: numerically(table), graphically, or alegbraically
The limit can still exist even though the limit could be undefined
If the limit from the left differs from the limit on the right, then the limit doesn’t exist
Approaching a limit from the positive side is written like this x–> n+ (let n be a number)
Approaching a limit from the negative side is written like this x–> n- (let n be a number)
Limits that fail to exist
1)Limits with jump discontinuity (any absolute value rational function)
2) Limits with infinite discontinuity (the limit approches oo or -oo)
3) The limit has ossilating(up and down rapidly) behavior sin(1/x)</p>
<p>Formal definition of a limit
lim f(x) = L
x–> c c is your x and L is an output
Delta epsilon proofs
|f(x) - L| < epsilon
So f(x) becomes arbitraly close to L. The above notation represents the distance between f(x) and L. The distance is your epsilon.
example
lim |2x-5| =1
x–> 3
|(2x-5)-1| < epsilon –> |2x-6|<epsilon --=“”> 2|x-3|<epsilon —=“”> |x-3| < (epsilon/2)
(epsilon/2) is your new delta
0 < |x -c| < delta
When we say “x appraches c” means that ther exists a positive number delta that is either (c - delta,c) or (c+delta, c)
so now the proof ends with 0 < |x-3| < (epsilon/2)</epsilon></epsilon></p>
<p>Ways to solve limits:
basic rules:
lim b = b limit x=c lim x^n=c^n (let n be a real number)
x–> c x–>c x–> c
direct substitution</p>
<p>properties of limits:
scalar multiple lim x –> c b(f(x)) = scalar *limit
sum or diff lim x–> c {f(x) ± g(x)] = limit + or - limit
product lim x–> c [f(x)g(x)] limit *limit
quotient lim x–> c f(x)/g(x) = limit/limit provided that g(x) doesn’t equal 0
power limt x–> c [f(x)]^n = limit^n</p>
<p>when direct sub fails hence you get (0/0)
factor (remember difference of squares, sum of cubes, and difference of cubes)
when dealing with trig functions, use the squeeze theorem
sinx/x =1 because of squeeze theorem proof
when you don’t have that directly multiply sinx/x by a fancy 1 for example
(3sinx/x) multiply that by (3/3) <–fancy 1 then you get (3<em>3sinx/3x) = 3</em>1 = 3
also when dealing with cos x or sinx (without sinx/x over x) use the squeeze thorem</p>
<p>For example:
lim (3 +cos x)/ (x+2)
x–>oo
-1 <= cos x <=1 (range of cosine)
+3 +3 +3 (the whole point is to make the function seen at the top and find what the function is between)
(2<=cos x +3 <= 4) / (x+2)
2/(x+2) <= (cos x +3)/(x+2) <= 4/(x+2) (now that i have made that function, make it approach oo since the original limit problem is approaching oo) Make the function that we just made h(x)
lim (2/(x+2)) <= h(x) <= lim 4/(x+2)
x–oo x–>oo
0<= h(x) <= 0
so lim (3 +cos x)/ (x+2) = 0
x–>oo</p>
<p>continous function - you can draw it without picking up your pencil
discontinous function - the opposite of what i just said</p>
<p>A function is continous if the 3 conditons are met
1)f(c) is defined
2)lim f(x) exits
x–> c
3) lim f(x) = f(c)
x–> c
If it doesn’t met all 3 then it is discontinous</p>
<p>3 types of discontinouties
- jump discontinuity - (example: |x|/x)
- removable discontinuity - you have a function that is undefined at that x value then there is a random dot at that x value that is not connected to the graph(however the limit still exists for this function. It would be the spot that is undefined)
- Infinite discontinuity - limit approaches oo or-oo</p>
<p>Intermediate Value Theorem(a common sense theroem) - if a function is continous on from [a,b] and k is an output number between f(a) and f(b) then there is at least 1 number c in [a,b] that makes </p>
<p>infinite limits (limits approaching oo)
if the degree of the numerator > degree of denominator then the limit is undefined
if degree of numerator = degree of denominator then take the ratio of the the highest degree’s coefficents
if the degree of the numerator < degree of denominator then the limit is 0
f(c) = k</p>