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Hyroko

  • one year ago

Use continuity to evaluate the limit: Sin^-1( sqrt(x^4+1)/(2x^2+3)) What does it mean by using continuity? If I were to write work for this on a test, how would I show my work?

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  1. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}{~~}\sin^{-1}\left(\frac{\sqrt{x^4+1}}{2x^2+3}\right)}\) like this ??

  2. hyroko
    • one year ago
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    correct I didn't know the syntax for writing it out like that.

  3. hyroko
    • one year ago
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    I know the answer is pi/6, I just need to know how to show my work using continuity.

  4. SolomonZelman
    • one year ago
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    using continuity, basically enables you to bring the limit inside.... (like I did below) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}{~~}\sin^{-1}\left(\frac{\sqrt{x^4+1}}{2x^2+3}\right)=\sin^{-1}\left(\lim_{x \rightarrow ~\infty}\frac{\sqrt{x^4+1}}{2x^2+3}\right)}\)

  5. hyroko
    • one year ago
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    because the limit of arcsine is continuous right?

  6. SolomonZelman
    • one year ago
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    because arcsine is a continuous (as well as differentiable) function.

  7. hyroko
    • one year ago
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    alright, I think I get it now have a calc exam today and had to make sure I was understanding everything Thank you for your help

  8. SolomonZelman
    • one year ago
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    Then, you can do this: \(\large\color{slate}{\displaystyle\sin^{-1}\left(\lim_{x \rightarrow ~\infty}\frac{\sqrt{x^4+1}}{2x^2+3}\right)}\) \(\large\color{slate}{\displaystyle\sin^{-1}\left(\lim_{x \rightarrow ~\infty}\frac{\sqrt{x^4+1}}{\sqrt{\left(2x^2+3\right)^2}}\right)}\) for infinitely large values 2x^2+3 is same as |2x^2+3|, so this is what allows me to perform to perform the step above. \(\large\color{slate}{\displaystyle\sin^{-1}\left(\lim_{x \rightarrow ~\infty}\sqrt{\frac{x^4+1}{\left(2x^2+3\right)^2}}\right)}\) then once again, you can use the fact that the function (the argument, the limit of which you are taking) is continuous. So you can bring the limit into the square root. \(\large\color{slate}{\displaystyle\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{x^4+1}{\left(2x^2+3\right)^2}}\right)}\)

  9. SolomonZelman
    • one year ago
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    when you expand this, you will get: \(\large\color{slate}{\displaystyle\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{x^4+1}{4x^4+12x^2+9}}~\right)}\) then, from there you know: \(\large\color{slate}{\displaystyle\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{1x^4+1}{4x^4+12x^2+9}}~\right)=\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{1}{4}}~\right)}\)

  10. SolomonZelman
    • one year ago
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    *√(1/4)*=1/2 So, the inverse sin of 1/2 is .....

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