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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?
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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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}{~~}\sin^{1}\left(\frac{\sqrt{x^4+1}}{2x^2+3}\right)}\) like this ??

hyroko
 one year ago
Best ResponseYou've already chosen the best response.0correct I didn't know the syntax for writing it out like that.

hyroko
 one year ago
Best ResponseYou've already chosen the best response.0I know the answer is pi/6, I just need to know how to show my work using continuity.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1using 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
 one year ago
Best ResponseYou've already chosen the best response.0because the limit of arcsine is continuous right?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1because arcsine is a continuous (as well as differentiable) function.

hyroko
 one year ago
Best ResponseYou've already chosen the best response.0alright, 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
 one year ago
Best ResponseYou've already chosen the best response.1Then, 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
 one year ago
Best ResponseYou've already chosen the best response.1when 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
 one year ago
Best ResponseYou've already chosen the best response.1*√(1/4)*=1/2 So, the inverse sin of 1/2 is .....
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