Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

using continuity, basically enables you to bring the limit inside....
(like I did below)

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

Answer 5

because the limit of arcsine is continuous right?

Answer 6

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

Answer 7

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

Answer 8

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.

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

Answer 9

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:

\(\large\color{slate}{\displaystyle\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{1x^4+1}{4x^4+12x^2+9}}~\right)=\sin^{-1}\left(\sqrt{\lim_{x \rightarrow ~\infty}\frac{1}{4}}~\right)}\)

Answer 10

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