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anonymous

  • one year ago

Please help one question! Use graphs and tables to find the limit and identify any vertical asymptotes of the function: lim x->2, 1/((x-2)^2) I know that it is +infinite and the asymptote is 2 but i dont know how to explain it

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  1. Luigi0210
    • one year ago
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    I guess you could say that as you approach 2 from the left it shoots up to infinity. And being 2 would leave you with an undefined value. The limit coming from the right to 2 would also approach infinity.

  2. anonymous
    • one year ago
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    But i dont know how to explain why it does that. Like I need to explain how I got my answer but I dont know how to explain it.

  3. anonymous
    • one year ago
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    @Luigi0210

  4. Luigi0210
    • one year ago
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    You could do what the instructions said and use a graph and table. Visually speaking, you can see it starts shooting up as it approaches 2. Mathematically, if you look at a table, the values start increasing as well. \(\large \lim_{x\rightarrow1.9} f(x)=100\) \(\large \lim_{x\rightarrow1.99} f(x)=10000\) \(\large \lim_{x\rightarrow1.999} f(x)=1000000\) \(\large \lim_{x\rightarrow1.99999} f(x)=+\infty\)

  5. anonymous
    • one year ago
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    I understand that part thank you, but what part of the equation shows the graph going up? I know that x -> 2 shows us the vertical asymptote of 2, but how does 1/((x-2)^2) show it going up?

  6. Luigi0210
    • one year ago
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    Are you asking how to clarify it is increasing?

  7. anonymous
    • one year ago
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    Yes. I

  8. Luigi0210
    • one year ago
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    plug in values? x=1, \(f(x)=\frac{1}{(x-2)^2} ===> f(1)\frac{1}{(1-2)^2}=\frac{1}{-1^2}=1\) x=1.5 \(f(x)=\frac{1}{(x-2)^2} ===> f(1.5)\frac{1}{(1.5-2)^2}=\frac{1}{-.5^2}=4\) x=1.75 \(f(x)=\frac{1}{(x-2)^2} ===> f(1.75)\frac{1}{(1.75-2)^2}=\frac{1}{-.25^2}=16\) x=1.9 \(f(x)=\frac{1}{(x-2)^2} ===> f(1)\frac{1.9}{(1.9-2)^2}=\frac{1}{-.1^2}=100\)

  9. Luigi0210
    • one year ago
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    *f(1.9) not f(1)

  10. anonymous
    • one year ago
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    Refer to the attached plot.

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  11. anonymous
    • one year ago
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    Thank you!

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