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Math Help - needing help on evalutating this limit..

  1. #1
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    needing help on evalutating this limit..

    hello, sorry for using my first post on this help forum ^^; I've been trying to ask this question everywhere on the net and no one seems to be able to help. So I'm trying here. Anyway

    I'm trying to find the limit of x tan (8/x) as x approaches to infinity, using the l'hospital's rule.

    So far I've applied the rule, but I'm stuck at the last part. Here's my work so far.

    Using the l'hospital's rule it turned out like this:
    sec^2 (8/x) * -8/x^2 / (-8/x^2)

    +the   -8/x^2's cancel out leaving me with
    <br />
sec^2 (8/x)

    so now this is where I'm stuck...When you input infinity in (8/x) , I can't seem to find where the values are heading to... for some examples

    8/80 = .1

    8/90 = 0.08888~

    8/100

    so.. would these values be heading towards 0 ? For my final answer I ended up getting, sec^2 (0) = 1 ... I think I made a mistake around here somehwere, but I don't know what.

    if you can help, it would be greatly appreciated!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by endiv View Post
    hello, sorry for using my first post on this help forum ^^; I've been trying to ask this question everywhere on the net and no one seems to be able to help. So I'm trying here. Anyway

    I'm trying to find the limit of x tan (8/x) as x approaches to infinity, using the l'hospital's rule.

    So far I've applied the rule, but I'm stuck at the last part. Here's my work so far.

    Using the l'hospital's rule it turned out like this:
    sec^2 (8/x) * -8/x^2 / -8/x^2

    +the  -8/x^2's cancel out leaving me with
    <br />
sec^2 (8/x)

    so now this is where I'm stuck...When you input infinity in (8/x) , I can't seem to find where the values are heading to... for some examples

    8/80 = .1

    [tex]8/90 = 0.08888~

    8/100

    so.. would these values be heading towards 0 ? For my final answer I ended up getting, sec^2 (0) = 1 ... I think I made a mistake around here somehwere, but I don't know what.

    if you can help, it would be greatly appreciated!
    If you have to apply the L'hopital's rule make it easier on yourself

    \lim_{n\to\infty}\frac{\tan\left(\frac{1}{n}\right  )}{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{\cos\lef  t(\frac{1}{n}\right)}\cdot\lim_{n\to\infty}\frac{\  sin\left(\frac{1}{n}\right)}{\frac{1}{n}}=\lim_{n\  to\infty}\frac{\sin\left(\frac{1}{n}\right)}{\frac  {1}{n}}

    Differentiating top and bottom givse

    \lim_{n\to\infty}\frac{\cos\left(\frac{1}{n}\right  )\cdot\frac{-1}{n^2}}{\frac{-1}{n^2}}=\lim_{n\to\infty}\cos\left(\frac{1}{n}\ri  ght)=1
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  3. #3
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    Ah, I found it ! I was confused comparing sin 8/x to the sin 1/x in your example. I thought that plugging the infinite in the 8/x would give me the limit so I ended up finding all this wierd answers. Thanks again!
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