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Math Help - Limits

  1. #1
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    Limits

    Can't remember how to do this (without L'Hopitals)

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

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

    lim x->0 (tan(6x))/(sin(2x))

    thanks for the help
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  2. #2
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    Quote Originally Posted by drain View Post
    Can't remember how to do this (without L'Hopitals)

    lim t->0 of (t^(3))/[tan(2t)]^(3)
    Heir.
    Attached Thumbnails Attached Thumbnails Limits-picture12.gif  
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by drain View Post
    Can't remember how to do this (without L'Hopitals)

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

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

    lim x->0 (tan(6x))/(sin(2x))

    thanks for the help
    I think for each of these we just need to rewrite the limits in a form that we can do:

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

    lim [t>0] [t^3/(sin(2t))^3]*(cos(2t))^3

    We can rewrite this as two limits multiplied together:

    lim [t>0] t^3/(sin(2t))^3 * lim [t>0] (cos(2t))^3

    lim [t>0] t^3/(sin(2t))^3 * (1)

    Now, I'll use the identity: sin(2t) = 2sint*cost

    lim [t>0] t^3/(2sint*cost)^3

    Rewritting this as the multiplication of limits again and then simplifying:

    lim [t>0] t^3/(2sint)^3 * (1)

    I'll factor the 1/2^3, and then rewrite the problem as:

    1/8 lim [t>0] (t/sint)^3 = 1/8 ... Since lim [t>0] of t/sint = 1

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

    I can't think of what to do for this one. I'll have to think about it some more and maybe get back to you with an answer.

    I encourage someone else to take a stab at this one!

    lim x->0 (tan(6x))/(sin(2x))

    This one we will treat similar to how I did the first. Any time I get multiplication or division of a cosine, I'm going to break it away and make it equal 1.

    lim [x>0] (sin(6x)/cos(6x))/(sin(2x))

    lim [x>0] (2sin(3x)cos(3x))/(2sin(x)cos(x)) <-- remove both cosines and reduce the 2/2.

    lim [x>0] sin(3x)/sin(x) <-- Note: sin(3x) = sin(x)cos(2x) + sin(2x)cos(x)

    lim [x>0] (sin(x)cos(2x) + sin(2x)cos(x))/sin(x) <-- separate this into an addition of limits

    lim [x>0] sin(x)cos(2x)/sin(x) + lim [x>0] sin(2x)cos(x)/sin(x) <-- remove both cosines

    1 + lim [x>0] 2sin(x)cos(x)/sin(x) <-- simplify

    1+2 = 3
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  4. #4
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    Thanks you guys. Greatly appreciated
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    I think for each of these we just need to rewrite the limits in a form that we can do:

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

    lim [t>0] [t^3/(sin(2t))^3]*(cos(2t))^3

    We can rewrite this as two limits multiplied together:

    lim [t>0] t^3/(sin(2t))^3 * lim [t>0] (cos(2t))^3

    lim [t>0] t^3/(sin(2t))^3 * (1)

    Now, I'll use the identity: sin(2t) = 2sint*cost

    lim [t>0] t^3/(2sint*cost)^3

    Rewritting this as the multiplication of limits again and then simplifying:

    lim [t>0] t^3/(2sint)^3 * (1)

    I'll factor the 1/2^3, and then rewrite the problem as:

    1/8 lim [t>0] (t/sint)^3 = 1/8 ... Since lim [t>0] of t/sint = 1

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

    I can't think of what to do for this one. I'll have to think about it some more and maybe get back to you with an answer.

    I encourage someone else to take a stab at this one!

    lim x->0 (tan(6x))/(sin(2x))

    This one we will treat similar to how I did the first. Any time I get multiplication or division of a cosine, I'm going to break it away and make it equal 1.

    lim [x>0] (sin(6x)/cos(6x))/(sin(2x))

    lim [x>0] (2sin(3x)cos(3x))/(2sin(x)cos(x)) <-- remove both cosines and reduce the 2/2.

    lim [x>0] sin(3x)/sin(x) <-- Note: sin(3x) = sin(x)cos(2x) + sin(2x)cos(x)

    lim [x>0] (sin(x)cos(2x) + sin(2x)cos(x))/sin(x) <-- separate this into an addition of limits

    lim [x>0] sin(x)cos(2x)/sin(x) + lim [x>0] sin(2x)cos(x)/sin(x) <-- remove both cosines

    1 + lim [x>0] 2sin(x)cos(x)/sin(x) <-- simplify

    1+2 = 3
    This is really hard to follow at first glance. I need to work on organizing my examples better
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  6. #6
    Junior Member frenzy's Avatar
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    Quote Originally Posted by drain View Post
    Can't remember how to do this (without L'Hopitals)

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

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

    lim x->0 (tan(6x))/(sin(2x))

    thanks for the help
    if latex was working i'd write up the 2nd one. It's not so I wont.

    i'll will get you started though


    multiply [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3)) by

    [sqrt(1+tan(t)) + sqrt(1+sin(t))]/[sqrt(1+tan(t)) + sqrt(1+sin(t))]]


    expand the numerator and then (after a little more work simplifying) try and find the limit.
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