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Math Help - Evaluating Limits Analyticall

  1. #1
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    Exclamation Evaluating Limits Analyticall

    Determine the limit of the trigonometric function (if it exists).

    1. lim sin x / 5x
    (x -> 0)
    <br />
    " alt="lim sin x / 5x
    (x -> 0)
    " />

    2. lim tan^2x / x
    (x ->0)

    3. lim cos x tan x / x
    (x -> 0)



    Below, find lim f(x+h) - f(x) / h
    ----------- (x->0)

    1. f(x) = radical x



    THANK YOU
    Last edited by Amadeus; October 1st 2007 at 07:44 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Amadeus View Post
    no one? T_T
    be patient
    Quote Originally Posted by Amadeus View Post
    Determine the limit of the trigonometric function (if it exists).

    1. lim sin x / 5x
    (x -> 0)
    lim       sin x / 5x   <br />
 (x -> 0)
    Note that: \lim_{x \to 0} \frac {\sin x}{5x} = \frac 15 \lim_{x \to 0} \frac {\sin x}x

    you should know what \lim_{x \to 0} \frac {\sin x}x is, so continue.

    2. lim tan^2x / x
    (x ->0)
    multiply the top and bottom by x:

    we get: \lim_{x \to 0} \frac {x \tan^2 x}{x^2} = \lim_{x \to 0} x  \left( \frac {\tan x}{x} \right)^2

    Now, \lim_{x \to 0} \frac {\tan x}x = \lim_{x \to 0} \frac {\sin x}x, so you should be able to finish this one too


    3. lim cos x tan x / x
    (x -> 0)
    \lim_{x \to 0} \cos x \left( \frac {\tan x}x\right) = \lim_{x \to 0}\cos x \cdot \lim_{x \to 0} \frac {\tan x}x

    same story as the last one. perhaps a nicer way to do it is to realize that \tan x = \frac {\sin x}{\cos x}, so the limit would simplify to: \lim_{x \to 0} \frac {\sin x}x

    ...i guess you're starting to see how important the limit \lim_{x \to 0} \frac {\sin x}x is, huh? ...you may want to simplify this way in the last problem

    Below, find lim f(x+h) - f(x) / h
    ----------- (x->0)

    1. f(x) = radical x
    You want: \lim_{h \to 0} \frac {\sqrt{x + h} - \sqrt {x}}h

    rationalize the numerator and simplify
    Last edited by Jhevon; October 2nd 2007 at 05:19 AM. Reason: fixed major typo
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    ^ Thank you so much, Jhevon! It really helped me a lot.

    But I'd like to check to see if my answers are right.

    1. 1/5
    2. 1
    3. 0 <-- I think that one is wrong...

    --

    1. 1 / radical h
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  4. #4
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    Quote Originally Posted by Amadeus View Post
    ^ Thank you so much, Jhevon! It really helped me a lot.

    But I'd like to check to see if my answers are right.

    1. 1/5
    yes

    2. 1
    No, for small x\ \tan(x) \approx x so (\tan(x))^2/x \approx x, so the limit is 0

    3. 0 <-- I think that one is wrong...
    Again for small x\ \cos(x) \approx 1\ \tan(x) \approx x so [\cos(x) \tan(x)]/x \approx 1
    the limit is 1

    RonL
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Amadeus View Post
    ^ Thank you so much, Jhevon! It really helped me a lot.

    But I'd like to check to see if my answers are right.

    1. 1/5
    CaptainBlack already told you this was correct

    2. 1
    remember, we were multiplying by an x here, and the x went to zero, so the second limit went to 1 and we multiplied by zero

    3. 0 <-- I think that one is wrong...
    here, you seem to have caught the fact that the x made us multiply by zero, however, the x was a typo (sorry) it should have been cos(x). (you should have caught that though, and i also mentioned that the limit simplified to sin(x)/x, so that's a second thing you should have caught)

    1. 1 / radical h
    you should get \frac 1{2 \sqrt {x}} (h is going to zero, there should be no h in your final answer, did you try my suggestion?)
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  6. #6
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    Thank you very much! I understand (for the most part)

    Although I got stuck on this one...

    Below, find lim f(x+h) - f(x) / h
    ----------- (x->0)


    f(x) = 4/x

    I tried it and got the first step as:

    (4/x+h) - (4/x) / h

    At this step, I'm pretty sure I'm on the wrong track but I multiplied both top & bottom by (4/x + h) + (4/x)

    Again, it turns out weird and I'm lost...
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  7. #7
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    Quote Originally Posted by Amadeus View Post
    Thank you very much! I understand (for the most part)

    Although I got stuck on this one...

    Below, find lim f(x+h) - f(x) / h
    ----------- (x->0)


    f(x) = 4/x

    I tried it and got the first step as:

    (4/x+h) - (4/x) / h

    At this step, I'm pretty sure I'm on the wrong track but I multiplied both top & bottom by (4/x + h) + (4/x)

    Again, it turns out weird and I'm lost...
    Your limit has a typo. It should be as h goes to 0.

    This one simply has a lot of algebra involved:
    \lim_{h \to 0}\frac{\frac{4}{x + h} - \frac{4}{x}}{h}

    Subtract the fractions:
    = \lim_{h \to 0}\frac{\frac{4x}{x(x + h)} - \frac{4(x + h)}{x(x + h)}}{h}

    = \lim_{h \to 0}\frac{4x - 4(x +h)}{hx(x + h)}

    = \lim_{h \to 0}\frac{-4h}{hx(x + h)}

    = \lim_{h \to 0}\frac{-4}{x(x + h)}

    Now take the limit:
    = \frac{-4}{x^2}

    -Dan
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