Hyroko
  • Hyroko
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?
Mathematics
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schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}{~~}\sin^{-1}\left(\frac{\sqrt{x^4+1}}{2x^2+3}\right)}\) like this ??
Hyroko
  • Hyroko
correct I didn't know the syntax for writing it out like that.
Hyroko
  • Hyroko
I know the answer is pi/6, I just need to know how to show my work using continuity.

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SolomonZelman
  • SolomonZelman
using continuity, basically enables you to bring the limit inside.... (like I did below) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}{~~}\sin^{-1}\left(\frac{\sqrt{x^4+1}}{2x^2+3}\right)=\sin^{-1}\left(\lim_{x \rightarrow ~\infty}\frac{\sqrt{x^4+1}}{2x^2+3}\right)}\)
Hyroko
  • Hyroko
because the limit of arcsine is continuous right?
SolomonZelman
  • SolomonZelman
because arcsine is a continuous (as well as differentiable) function.
Hyroko
  • Hyroko
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
SolomonZelman
  • SolomonZelman
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. \(\large\color{slate}{\displaystyle\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{x^4+1}{\left(2x^2+3\right)^2}}\right)}\)
SolomonZelman
  • SolomonZelman
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: \(\large\color{slate}{\displaystyle\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{1x^4+1}{4x^4+12x^2+9}}~\right)=\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{1}{4}}~\right)}\)
SolomonZelman
  • SolomonZelman
*√(1/4)*=1/2 So, the inverse sin of 1/2 is .....

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