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eva12

  • one year ago

1)test review let f(x)=1/sqrt(x) and 0<c<1. Then f is continuous at x=c.proof

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  1. eva12
    • one year ago
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    definition 55.a function f is said to be continuous at a point x if for every epsilon >0 there exists \[\delta>0\]such that |f(y)-f(x)| < epsilon for all y such that |y-x| \[<\delta\]

  2. zepdrix
    • one year ago
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    For \(\large\rm f(x)=\frac{1}{\sqrt{x}}\) to be continuous at c means \(\large\rm \forall\epsilon>0~~ \exists\delta>0:(x\in(0,1),|x-c|<\delta)\implies \color{orangered}{|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{c}}|<\epsilon}\) Let's fiddle around with that orange part a little bit, maybe we can get somewhere.

  3. zepdrix
    • one year ago
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    Getting a common denominator gives us: \(\large\rm \left\lvert\dfrac{\sqrt{c}-\sqrt{x}}{\sqrt{cx}}\right\rvert\) since we're in absolute value, this is equivalent to\(\large\rm \dfrac{1}{\left\lvert\sqrt{cx}\right\rvert}\left\lvert\sqrt x-\sqrt c\right\rvert\) which is still less than our \(\large\rm \epsilon\).\[\large\rm \dfrac{1}{\left\lvert\sqrt{cx}\right\rvert}\left\lvert\sqrt x-\sqrt c\right\rvert<\epsilon\]Multiply the sqrt(cx) to the other side, and multiply both sides by the conjugate of what's in the absolute,\[\large\rm \left\lvert\sqrt x-\sqrt c\right\rvert\left\lvert\sqrt x+\sqrt c\right\rvert<\epsilon\left\lvert\sqrt x+\sqrt c\right\rvert\left\lvert\sqrt{cx}\right\rvert\]\[\large\rm \left\lvert x-c\right\rvert<\epsilon\left\lvert\sqrt x+\sqrt c\right\rvert\left\lvert\sqrt{cx}\right\rvert\]

  4. zepdrix
    • one year ago
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    I think we can drop the absolutes on the right safely... since it's all addition and square roots,\[\large\rm \left\lvert x-c\right\rvert<\epsilon\left(\sqrt x+\sqrt c\right)\sqrt{cx}\]

  5. zepdrix
    • one year ago
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    Actually, lemme color these a sec before I go any further,\[\large\rm \left\lvert x-c\right\rvert<\epsilon\color{royalblue}{\left(\sqrt x+\sqrt c\right)}\color{green}{\sqrt{cx}}\] So we were able to, sort of, build our delta inequality out of the epsilon one. So for \(\large\rm \delta<1\) we have:\[\large\rm |x-c|<1\]\[\large\rm -1<x-c<1\]\(\large\rm c-1<x<c+1\) Let's call this equation (1).

  6. zepdrix
    • one year ago
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    Multiplying equation (1) by c, and then taking the square root gives us,\[\large\rm \sqrt{c^2-c}<\sqrt{cx}<\sqrt{c^2+c}\]This helps us get rid of this dependence on x here,\[\large\rm \sqrt{c^2-c}<\color{green}{\sqrt{cx}<\sqrt{c^2+c}}\]

  7. zepdrix
    • one year ago
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    Going back, taking the square root of equation (1), and then adding sqrt(c) to all sides,\[\large\rm \sqrt{c-1}+\sqrt c<\sqrt{x}+\sqrt{c}<\sqrt{c+1}+\sqrt{c}\]This also helps us to get rid of that pesky x!\[\large\rm \sqrt{c-1}+\sqrt c<\color{royalblue}{\sqrt{x}+\sqrt{c}<\sqrt{c+1}+\sqrt{c}}\]

  8. zepdrix
    • one year ago
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    So we have that when \(\large\rm \delta<1\), \[\large\rm \left\lvert x-c\right\rvert\quad<\quad\epsilon\color{royalblue}{\left(\sqrt x+\sqrt c\right)}\color{green}{\sqrt{cx}}\quad<\quad\epsilon\color{royalblue}{\left(\sqrt{c+1}+\sqrt{c}\right)}\color{green}{\sqrt{c^2+c}}\]

  9. zepdrix
    • one year ago
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    \[\large\rm |x-c|<\epsilon(\sqrt{c+1}+\sqrt{c})\sqrt{c^2+c}\]

  10. zepdrix
    • one year ago
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    Let \(\large\rm \delta=\min\left\{1,~~\epsilon(\sqrt{c+1}+\sqrt{c})\sqrt{c^2+c}\right\}\)

  11. zepdrix
    • one year ago
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    I'm not really sure what we would say about \(\large\rm \delta>1\) though, hmm... Hopefully I did something right in there though D: Sorry I'm kinda new at these types of problems, just trying to see if I could work through it properly. I think that proves it for \(\large\rm \delta<1\).

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