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Math Help - Some Calculus limits

  1. #1
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    Unhappy Some Calculus limits

    I can't for the life of me remember how to do these

    lim t->0 of (t^(3))/[tan(2t)]^(3)

    lim t->0 of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))



    These 2 also =(

    evaluate dy if y=x^(3) - 2x^(2) + 1 if x=2 and dx= 0.2 I don't see why you would need dx? Wouldn't you just differentiate x and then plug in the 2? hmm...

    And this last one

    approximate sqrt(64.05) using differentials.

    THANKS IN ADVANCE!!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by drain View Post
    lim t->0 of (t^(3))/[tan(2t)]^(3)
    obviously we have a problem. the top and bottom go to zero. since we have a 0/0 as we take the limit, we can use L'Hoptial's rule (do you remember it?)

    to make life easier, i will also use a limit theorem here that says lim A*B = (lim A)*(lim B)

    so lim t->0 of (t^(3))/[tan(2t)]^(3)

    Apply L'hopital's, we get

    lim t->0 of (3t^2)/[3(tan(2t))^2 * 2sec^2(2t)]

    we still get 0/0 as we take the limit, we will apply l'hopital's again. but before i do, i want to split up this function, so i don't have to find the derivative of [3(tan(2t))^2 * 2sec^2(2t)]

    notice that (3t^2)/[3(tan(2t))^2 * 2sec^2(2t)] = (3t^2)/[3(tan(2t))^2] *1/( 2sec^2(2t))

    so lim t->0 of (3t^2)/[3(tan(2t))^2 * 2sec^2(2t)]
    = lim t->0 of (3t^2)/[3(tan(2t))^2] * 1/(2sec^2(2t))
    = lim t->0 of (3t^2)/[3(tan(2t))^2] * lim t->0 1/(2sec^2(2t))

    now the second limit is okay, we will do l'hopital's on the first, we get:

    = lim t->0 of (6t)/[6(tan(2t)) * 2sec^2(2t)] * lim t->0 1/(2sec^2(2t))

    we still have a problem with the first limit. i split it again and perform l'hopital's on the first piece

    lim t->0 of (6t)/[6(tan(2t)) * 2sec^2(2t)] * lim t->0 1/(2sec^2(2t))
    = lim t->0 of (6t)/[6(tan(2t))] * lim t->0 1/(2sec^2(2t)) * lim t->0 1/(2sec^2(2t))

    apply l'hopital's to the first:

    = lim t->0 of (6)/[6*2sec^2(2t)] * lim t->0 1/(2sec^2(2t)) * lim t->0 1/(2sec^2(2t))

    now we can plug in. notice that sec^2(2t) = 1/cos^2(2t) and cos(0) = 1

    taking the limits we obtain:

    = 1/2 * 1/2 * 1/2
    = 1/8



    lim t->0 of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))
    the first term is ok for taking limits, but the second piece, sqrt(1+sin(t))]/(t^(3)) has a problem. the limit makes the denominator zero, and we can't have that. however, we can't use l'hoptial's on this one since we don't have 0/0 or infinity/infinity, and thank God for that, since differentiating that 3 times over will be a pain. so what do we do.

    notice that we have 1/0 as we take the limit, so it would hint that the limit for this piece is infinity, but not so fast, we have to make sure of that. if we approach zero from the left the limit is -infinity, if we approach it from the right, the limit is +infinity (do you see that?). since the right hand limit is not equal to the left hand limit, the limit does not exist. so the limit of this whole function does not exist.



    evaluate dy if y=x^(3) - 2x^(2) + 1 if x=2 and dx= 0.2
    y = x^3 - 2x^2 + 1
    => dy/dx = 3x^2 - 4x
    => dy = dx(3x^2 - 4x) ............i multiplied through by dx
    when dx = 0.2 and x = 2 we get
    dy = (0.2)(3(2)^2 - 4(2)
    => dy = 0.8



    approximate sqrt(64.05) using differentials.
    we will use linear approximation using differentials to do this problem.

    recall that f(x) ~= f(a) + f ' (a)(x - a)

    x is what we want to know, a is something close to x that we already know. Huh??? let me explain.

    we want to know sqrt(64.05). if we write this as a function we can write it as f(x) = sqrt(x), and we want to know f(64.05).

    so what is f(64.05). i dunno, but i sure as heck know what f(64) is. sqrt(64) = 8. and 64 is close to 64.05.

    so now we use x = 64.05 and a = 64, let's do the problem.


    Consider the function f(x) = sqrt(x)

    now f ' (x) = 1/2*sqrt(x)

    using f(x) ~= f(a) + f ' (a)(x - a) with x = 64.05 and a = 64, we will approximate the value of f(64.05), that is sqrt(64.05)

    Now f(64.05) ~= f(64) + f ' (64)(64.05 - 64)
    => f(64.05) ~= sqrt(64) + 1/2sqrt(64) * (0.05)
    => f(64.05) ~= 8 + 0.0625 * (0.05)
    => f(64.05) ~= 8.003125

    so sqrt(64.05) ~= 8.003125 by the method of differentials.

    in case you curious, the actual value is 8.00312439, pretty close huh!
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  3. #3
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    Ah! Everything helped out except for the first two, because I have no idea how to do "L'Hopital's" or whatever. Never even heard of that Is there another way to go about doing those problems? Thanks for the help on the other two though. =)
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  4. #4
    Junior Member frenzy's Avatar
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    Quote Originally Posted by drain View Post
    ...

    lim t->0 of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))
    Hmm..i'm getting 1/4 for this one.
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  5. #5
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    Quote Originally Posted by frenzy View Post
    Hmm..i'm getting 1/4 for this one.
    I confirm
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  6. #6
    Junior Member frenzy's Avatar
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    you can do the first two problems without using L'hopital's rule if you use the fact


    lim as t->0 sin(t)/t=1
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by frenzy View Post
    Hmm..i'm getting 1/4 for this one.
    o sorry, i thought it was two separate functions!

    i thought i saw sqrt(1+tan(t)) - [sqrt(1+sin(t))]/(t^(3))

    instead of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))

    now i see it was one. my bad. you guys can give the solution if you want


    sorry about that drain, i must be tired or something
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  8. #8
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    Thanks to all of you! =) Really do appreciate it.
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