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Math Help - limit problem

  1. #1
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    limit problem

    I am supposed to find this limit without L'Hopital's rule. I have tried multiplying the numerator and the denominator with (1/(1-x))+1, and I've looked for ways to factorize, without getting anywere. Would appreciate any help! =)
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    MHF Contributor Also sprach Zarathustra's Avatar
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    lim \frac{\frac{1}{x+1} -1}{x}=lim \frac{-1}{x+1}

    so when x \to 0 the above equals to -1
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    Thank you very much! Would you mind explaining exactly how you simplified the limit? Did you just muliply the numerator and the denominator with (x+1)? I would never have spotted that, thanks again!
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Of course...

    lim \frac{\frac{1}{x+1} -1}{x}=lim \frac{\frac{1-1 \cdot (x+1)}{x+1}}{x}=lim \frac{\frac{1-x-1}{x+1}}{x}=lim \frac{\frac{-x}{x+1}}{x}=lim \frac{-x}{x(x+1)}=lim\frac{-1}{x+1}
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    MHF Contributor Also sprach Zarathustra's Avatar
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    And something more...

    I notice that you have 11^2 posts already and yet you haven't try write math with Latex, I strongly recommend you to try it!

    You can start learning this here, on this site there is a forum about Latex...

    ...but the best way I think is by double-clicking on the code, you can try it with my post above!


    Anyway... good-luck!
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    Another quick way to see this is to note that if f(x) = \frac{1}{1 + x} then

     \displaystyle<br />
\lim_{x \to 0} \frac{\frac{1}{1 + x} - 1}{x} = f'(0)<br />

    by definition. Given the way the limit was set up, this is probably what was expected.
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    Thanks alot! I tried using Latex for this next problem, but I kept getting errors. I will have to start using it on easier expressions first I think... Anyway, in this problem, is it possible for me to go from step 2 to step 3 like I did here? In that case, the rest of the problem is easy.
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    You can do this using the same trick I mentioned. If f(x) = \sin x then

    \displaystyle<br />
\lim_{h \to 0} \frac{f \left(\frac{\pi}{6} + h \right) - 1/2}{h} = f'\left(\frac{\pi}{6}\right) = \cos  \frac{\pi}{6}.<br />

    To get at the same thing you are suggesting here, note that you can rewrite the second expression as

     \displaystyle<br />
  \frac{1}{2}   \frac{(\cos x - 1)}{x}     + \cos \left(\frac{\pi}{6}\right) \frac{\sin x}{x}.<br />

    Then take the limit, using the usual results for this types of things.
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    That will indeed work! I've never seen anyone using the defintion of the derivative in solving limit problems like that, haha. But if I do it like I did, then it will also simplify to 1 times cos (pi/6) which is what you got. I'm just not sure if I can rewrite it like that. It would be nice to know for future problems.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by gralla55 View Post
    Thanks alot! I tried using Latex for this next problem, but I kept getting errors. I will have to start using it on easier expressions first I think... Anyway, in this problem, is it possible for me to go from step 2 to step 3 like I did here? In that case, the rest of the problem is easy.
    No! This is wrong...

    But you can solve this problem by definition of derivative:

    f'(x_0)=lim_{\Delta{x}\to 0} \frac{f(x+\Delta{x})-f(x_0)}{\Delta{x}}

    Hint:

    f(x) in your case is: f(x)=sin(x)}

    And you know that:
    sin(\frac{\pi}{6})=\frac{1}{2}
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    Quote Originally Posted by gralla55 View Post
    That will indeed work! I've never seen anyone using the defintion of the derivative in solving limit problems like that, haha. But if I do it like I did, then it will also simplify to 1 times cos (pi/6) which is what you got. I'm just not sure if I can rewrite it like that. It would be nice to know for future problems.
    Yeah, see the edit I made. It works it out explicitly. What you can do is, if \lim f(x) = A and \lim g(x) = B then the following hold:

    \displaystyle<br />
\lim(f(x) + g(x)) = A + B \qquad \lim f(x) g(x) = AB \qquad \lim \frac{f(x)}{g(x)} = \frac{A}{B}<br />

    provided, in the last case, that \lim g(x) \ne 0. You can't do what you did though, as a general rule.
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    Thank you both! Very helpful! Using the last limit properties, can I solve the problem like this?
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  13. #13
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    Quote Originally Posted by gralla55 View Post
    Thanks alot! I tried using Latex for this next problem, but I kept getting errors. I will have to start using it on easier expressions first I think... Anyway, in this problem, is it possible for me to go from step 2 to step 3 like I did here? In that case, the rest of the problem is easy.
    ANOTHER SOLUTION:

    Recall the trigonometric identity:

    sin(x)-sin(y)=2cos(\frac{x+y}{2})sin(\frac{x-y}{2})

    Now, as we know: \frac{1}{2}=sin(\frac{\pi}{6})

    Therefore:

    sin(\frac{\pi}{6}+x)+\frac{1}{2}=sin(\frac{\pi}{6}  +x)+sin(\frac{\pi}{6})=2cos(\frac{\pi}{6}+\frac{x}  {2})sin(\frac{x}{2})

    Now, you can continue this using:

    lim_{x\to 0} \frac{sin(x)}{x}=1
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  14. #14
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    Quote Originally Posted by gralla55 View Post
    Thank you both! Very helpful! Using the last limit properties, can I solve the problem like this?
    You not allowed to do that! (Some of the steps are incorrect!)
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  15. #15
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    You can't move the limit taking into the denominator, because \lim_{x \to 0} x = 0. The rule for moving limits around doesn't allow for this. The result would be undefined to begin with.

    The limit results you want to use (taken as given) are:

     \displaystyle<br />
\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{\cos x - 1}{x} = 0<br />
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