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Math Help - Proof of a limit

  1. #1
    Super Member redsoxfan325's Avatar
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    Proof of a limit

    Prove that \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{10} and from this deduce that \frac{1}{x^2+1} is continuous at x=3.

    Proposed solution:

    \left |\frac{1}{10} - \frac{1}{x^2+1}\right | = \left | \frac{x^2+1}{10x^2+10} - \frac{10}{10x^2+10}\right | = \frac{1}{10}\left |\frac{x^2-9}{x^2+1}\right | = \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right |

    If 0 < |x-3| < 1, then |x+3|<7 and x^2+1>10

    Thus, \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right | < \frac{1}{10}\left |\frac{7|x-3|}{10} \right | = \frac{7}{100}|x-3|.

    Thus, let \delta = \min\left\{1, \frac{100\epsilon}{7}\right\} therefore when |x-3|<\delta, \left |\frac{1}{10} - \frac{1}{x^2+1}\right |< \epsilon.

    Since \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{3^2+1} = \frac{1}{10}, we can conclude that \frac{1}{x^2+1} is continuous at x=3.

    Is that correct, and if not, how do I fix it?

    Now, what if I need to prove that \frac{1}{x^2+1} is continuous for all x. How would I find an appropriate delta then?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    Prove that \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{10} and from this deduce that \frac{1}{x^2+1} is continuous at x=3.

    Proposed solution:

    \left |\frac{1}{10} - \frac{1}{x^2+1}\right | = \left | \frac{x^2+1}{10x^2+10} - \frac{10}{10x^2+10}\right | = \frac{1}{10}\left |\frac{x^2-9}{x^2+1}\right | = \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right |

    If 0 < |x-3| < 1, then |x+3|<7 and x^2+1>10

    Thus, \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right | < \frac{1}{10}\left |\frac{7|x-3|}{10} \right | = \frac{7}{100}|x-3|.

    Thus, let \delta = \min\left\{1, \frac{100\epsilon}{7}\right\} therefore when |x-3|<\delta, \left |\frac{1}{10} - \frac{1}{x^2+1}\right |< \epsilon.
    very nice, but a little snag: x^2 + 1 > 10 is not always true with these conditions.

    Since \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{3^2+1} = \frac{1}{10}, we can conclude that \frac{1}{x^2+1} is continuous at x=3.

    Is that correct, and if not, how do I fix it?
    yes, it is fine to show that \lim_{x \to x_0}f(x) = f(x_0), this means f(x) is continuous at x_0

    Now, what if I need to prove that \frac{1}{x^2+1} is continuous for all x. How would I find an appropriate delta then?
    you need to show the definition holds for all points in the domain of the function (which happens to be all real numbers). so let x_0 be an arbitrary point in the domain. you must show:

    For every \epsilon > 0, there exists a \delta > 0 such that x,x_0 \in \text{dom}(f) and 0<|x - x_0| < \delta implies |f(x) - f(x_0)|< \epsilon

    to find your delta, you would use similar manipulations to what you have done so far
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  3. #3
    Super Member redsoxfan325's Avatar
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    OK, so I'll change 10 to 5.

    Because 0 < |x-3| < 1, then either 0<x-3<1, in which case 5 still works ( 5<10<x^2+1), or 0>x-3>-1, so x^2+1>5.

    Thus I'll let \delta = \min\left\{1, \frac{50\epsilon}{7}\right\}.
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    Prove that \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{10} and from this deduce that \frac{1}{x^2+1} is continuous at x=3.
    We need to show that:

    |\frac{1}{x^2+1}-\frac{1}{x_0^2+1}|<\epsilon

    Which simplifies to:

    \frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\epsilon

    But we have to get that beginning with delta:

    |x-x_0|<\delta

    So, let's set \delta\leq\frac{|x_0|}{2}

    Which gives us:

    |x-x_0|<\delta\leq\frac{|x_0|}{2}\Rightarrow|x-x_0|<\frac{|x_0|}{2}

    \Rightarrow-\frac{|x_0|}{2}<x-x_0<\frac{|x_0|}{2}

    \Rightarrow-\frac{5|x_0|}{2}\leq-\frac{|x_0|+4x_0}{2}<x+x_0<\frac{|x_0|+4x_0}{2}\le  q\frac{5|x_0|}{2}

    \Rightarrow|x+x_0|<\frac{5|x_0|}{2}

    \Rightarrow|x-x_0||x+x_0|<\frac{5|x_0|\delta}{2}

    Observe that: \frac{1}{(x_0^2+1)(x^2+1)}\leq1

    So: \frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}\leq|x-x_0||x+x_0|

    \Rightarrow\frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\frac{5|x_0|\delta}  {2}

    Now we can just choose \delta=\min\{\frac{x_0}{2},\frac{2\epsilon}{5|x_0|  }\}

    \Rightarrow \delta\leq\frac{2\epsilon}{5|x_0|}\Rightarrow\frac  {5|x_0|\delta}{2}\leq\epsilon

    \Rightarrow\frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\frac{5|x_0|\delta}  {2}\leq\epsilon

    \Rightarrow\frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\epsilon

    So \lim_{x\to x_0}\frac{1}{x^2+1}=\frac{1}{x_0^2+1}\;\forall\;x\  in\mathbb{R}

    Which means \frac{1}{x^2+1} is continuous for all real x.
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