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Math Help - Trouble with limits

  1. #1
    Senior Member Spec's Avatar
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    Trouble with limits

    I've encountered some difficult problems in my textbook that I can't solve. Any help would be appreciated. This is not homework—I'm just looking to advance my knowledge, that's all.

    Question 1: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
    For each \epsilon > 0 , find a \omega so that x > \omega \implies \left| {\frac{\sqrt{x^2 + 1}}{x} - 1 }\right| < \epsilon is true. What does this show?

    (By the way, how do I get the absolute characters to extend all the way in LaTeX?) SOLVED: Use \left| and \right|


    Question 2: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
    Show that x^3 \to 1 when x \to 1 using the definition of limits.

    This is what I've got so far:

    As soon as we've chosen a \epsilon > 0, then we'll be able to find a \delta > 0 so that
    |x^3 - 1| < \epsilon when 0 < |x - 1| < \delta.

    Note:
    |x^3 - 1| = |x - 1| \cdot |x^2 + x + 1|



    Question 3: (SOLVED)
    Is f(x) = \lim_{n \to \infty} \frac{x^n}{1 + x^n} continuous for all x \geq 0?

    Is f(x) = \lim_{n \to \infty} \frac{x + x^2 + ... + x^n}{1 + x + ... + x^n} continuous for all x \geq 0?
    Last edited by Spec; November 3rd 2007 at 05:27 PM. Reason: Fixed LaTeX code
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Spec View Post

    Question 2:
    Show that x^3 \to 1 when x \to 1 using the definition of limits.

    This is what I've got so far:

    As soon as we've chosen a \epsilon > 0, then we'll be able to find a \delta > 0 so that
    |x^3 - 1| < \epsilon when 0 < |x - 1| < \delta.

    Note:
    |x^3 - 1| = |x - 1| \cdot |x^2 + x + 1|
    if |x| < 1 \, or \, (-1 < x < 1) \implies 0 < x^2 < 1 so that

     -1 < x^2 + x < 2 \implies 0 < x^2 + x + 1 < 3

    \implies |x^3 - 1| = |x - 1| \cdot |x^2 + x + 1| < 3\delta
    then choose \delta = \frac{\epsilon}{3}
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Spec View Post
    Is f(x) = \lim_{n \to \infty} \frac{x^n}{1 + x^n} continuous for all x \geq 0?
    Consider x = 0. Now consider all other x. Can you define an epsilon and delta around x = 0?

    -Dan
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  4. #4
    MHF Contributor red_dog's Avatar
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    If x\in[0,1) then \lim_{n\to\infty}x^n=0\Rightarrow f(x)=0.
    If x=1\Rightarrow f(1)=\frac{1}{2}
    If \displaystyle x>1\Rightarrow\lim_{n\to\infty}\frac{x^n}{1+x^n}=\  lim_{n\to\infty}\frac{x^n}{x^n\left[\left(\frac{1}{x}\right)^n+1\right]}=1.
    So \displaystyle f(x)=\left\{\begin{array}{cc}0, & x\in[0,1)\\\frac{1}{2}, & x=1\\1, & x>1\end{array}\right., and f is discontinuous in x=1.


    For the second one:
    f(x)=\left\{\begin{array}{cc}x, & x\in[0,1)\\\frac{n}{n+1}, & x=1\\1, & x>1\end{array}\right. and f is discontinuous in x=1.
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  5. #5
    Senior Member Spec's Avatar
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    Quote Originally Posted by red_dog View Post
    If x\in[0,1) then \lim_{n\to\infty}x^n=0\Rightarrow f(x)=0.
    If x=1\Rightarrow f(1)=\frac{1}{2}
    If \displaystyle x>1\Rightarrow\lim_{n\to\infty}\frac{x^n}{1+x^n}=\  lim_{n\to\infty}\frac{x^n}{x^n\left[\left(\frac{1}{x}\right)^n+1\right]}=1.
    So \displaystyle f(x)=\left\{\begin{array}{cc}0, & x\in[0,1)\\\frac{1}{2}, & x=1\\1, & x>1\end{array}\right., and f is discontinuous in x=1.


    For the second one:
    f(x)=\left\{\begin{array}{cc}x, & x\in[0,1)\\\frac{n}{n+1}, & x=1\\1, & x>1\end{array}\right. and f is discontinuous in x=1.
    The second one is supposed to be continuous for all x \geq 0 according to the answer in the textbook.
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  6. #6
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    Quote Originally Posted by Spec View Post
    I've encountered some difficult problems in my textbook that I can't solve. Any help would be appreciated. This is not homework—I'm just looking to advance my knowledge, that's all.

    Question 1:
    For each \epsilon > 0 , find a \omega so that x > \omega \implies |{\frac{\sqrt{x^2 + 1}}{x} - 1|} < \epsilon is true. What does this show?
    This would show that,
    \lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x} = 1.

    Note that if x>0 then:
    \left| \frac{\sqrt{x^2+1}}{x} - 1 \right| = \left| \frac{\sqrt{x^2+1}-x}{x} \cdot \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}+x} \right|  = \frac{1}{x(\sqrt{x^2+1}+x)} \leq \frac{1}{x(\sqrt{x^2}+x)} = \frac{1}{2x^2} \leq \frac{1}{x^2}.
    So chose,
    C > \sqrt{\frac{1}{\epsilon}}.
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  7. #7
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    Quote Originally Posted by Spec View Post
    The second one is supposed to be continuous for all x \geq 0 according to the answer in the textbook.
    The book is wrong.
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  8. #8
    Senior Member Spec's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    This would show that,
    \lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x} = 1.

    Note that if x>0 then:
    \left| \frac{\sqrt{x^2+1}}{x} - 1 \right| = \left| \frac{\sqrt{x^2+1}-x}{x} \cdot \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}+x} \right|  = \frac{1}{x(\sqrt{x^2+1}+x)} \leq \frac{1}{x(\sqrt{x^2}+x)} = \frac{1}{2x^2} \leq \frac{1}{x^2}.
    So chose,
    C > \sqrt{\frac{1}{\epsilon}}.
    I have no idea what you did there. I'm with you right until this point (why did you choose to compare it to that?):

     \leq \frac{1}{x(\sqrt{x^2}+x)} = \frac{1}{2x^2} \leq \frac{1}{x^2}.
    So chose,
    C > \sqrt{\frac{1}{\epsilon}}.
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  9. #9
    MHF Contributor kalagota's Avatar
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    because you are sure that the latter function is greater than the absolute thing, and it easier to compute, if that latter function is less than epsilon, then your absolute thing is less than epsilon..
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  10. #10
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by Spec View Post
    The second one is supposed to be continuous for all x \geq 0 according to the answer in the textbook.
    Sorry, you're right. I forgot to take the limit in the case x=1.
    \displaystyle\lim_{n\to\infty}\frac{n}{n+1}=1.
    So, f(x)=\left\{\begin{array}{cc}x, & x\in[0,1)\\1, & x\geq 1\end{array}\right.
    and f is continuous for all x\geq 0
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  11. #11
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Spec View Post
    ...
    Question 1: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
    For each \epsilon > 0 , find a \omega so that x > \omega \implies \left| {\frac{\sqrt{x^2 + 1}}{x} - 1 }\right| < \epsilon is true. What does this show?

    (By the way, how do I get the absolute characters to extend all the way in LaTeX?) SOLVED: Use \left| and \right|


    Question 2: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
    Show that x^3 \to 1 when x \to 1 using the definition of limits.

    This is what I've got so far:

    As soon as we've chosen a \epsilon > 0, then we'll be able to find a \delta > 0 so that
    |x^3 - 1| < \epsilon when 0 < |x - 1| < \delta.

    Note:
    |x^3 - 1| = |x - 1| \cdot |x^2 + x + 1|
    ...

    you'll just get used to it.. Ü
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  12. #12
    Senior Member Spec's Avatar
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    I need help with another limit.


    \lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}}

    This is what I've done so far.

    \lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}} = \lim_{x \to \infty}\frac{\frac{x}{ln(e^{2x} + x)} + 1}{\frac{\sqrt{x}}{e^{1 + ln(x)}} + 1} \cdot \frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}

    If this is correct, then the question becomes; what can I do with:

    \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}



    Also, I've solved this one, but my solution seems a bit strange. Anyone have a better solution?

    \lim_{x \to \infty}(1 + \frac{1}{5x})^{7x}

    I used a substitution:
    t = \frac{1}{x}, t \to 0 when x \to \infty

    So we have:
    \lim_{t \to 0} (1 + \frac{t}{5})^{\frac{7}{t}} = \lim_{t \to 0} e^{ln(1 + \frac{t}{5})^{\frac{7}{t}}} = \lim_{t \to 0} e^{\frac{7}{5} \cdot ln(1 + \frac{t}{5})^{\frac{7}{t}^\frac{5}{7}}} = \lim_{t \to 0} e^{\frac{7}{5} \cdot ln(1 + \frac{t}{5})^{\frac{5}{t}}} = \lim_{t \to 0} e^{\frac{7}{5} \cdot ln(e)} = e^{\frac{7}{5}}
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Spec View Post
    I need help with another limit.


    \lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}}

    This is what I've done so far.

    \lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}} = \lim_{x \to \infty}\frac{\frac{x}{ln(e^{2x} + x)} + 1}{\frac{\sqrt{x}}{e^{1 + ln(x)}} + 1} \cdot \frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}

    If this is correct, then the question becomes; what can I do with:

    \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}
    Here's a possible route: Have you noticed that e^{1 + ln(x)} = ex? Then you can use ThePerfectHacker's favorite method: the squeeze theorem.

    -Dan
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  14. #14
    Senior Member Spec's Avatar
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    I'm not quite sure how to apply the squeeze theorem here. Are there any other solutions that doesn't require its use?
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  15. #15
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Spec View Post
    \lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}}

    This is what I've done so far.

    \lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}} = \lim_{x \to \infty}\frac{\frac{x}{ln(e^{2x} + x)} + 1}{\frac{\sqrt{x}}{e^{1 + ln(x)}} + 1} \cdot \frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}

    If this is correct, then the question becomes; what can I do with:

    \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}
    Sorry, I don't know what I was thinking about for an upper bound on the limit.

    What I would do is this:
    \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}} = \frac{1}{e} \cdot \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{x}

    This limit is of the form \frac{\infty}{\infty} so we may use L'Hopital's rule here:
    \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}} = \frac{1}{e} \cdot \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{x} = \frac{1}{e} \cdot \lim_{x \to \infty} \frac{\frac{1}{e^{2x} + x} \cdot (2e^{2x} + 1)}{1}  = \frac{1}{e} \cdot \lim_{x \to \infty} \frac{2e^{2x} + 1}{e^{2x} + x}

    It should be clear that the value of this last limit is 2.

    -Dan
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