## anonymous 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?

1. SolomonZelman

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

2. anonymous

correct I didn't know the syntax for writing it out like that.

3. anonymous

I know the answer is pi/6, I just need to know how to show my work using continuity.

4. SolomonZelman

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

5. anonymous

because the limit of arcsine is continuous right?

6. SolomonZelman

because arcsine is a continuous (as well as differentiable) function.

7. anonymous

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

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

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

*√(1/4)*=1/2 So, the inverse sin of 1/2 is .....