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Mathematics 8 Online
OpenStudy (anonymous):

find the limit as x approaches infinity (sqrt(t)+t^2)/(2t-t^2) it equals -1? can someone show me the steps?

OpenStudy (anonymous):

Split the fraction into sums of fractions: \[\lim_{t \rightarrow \infty}\frac{\sqrt{t} + t^2}{2t - t^{2}} = \lim_{t \rightarrow \infty}\frac{\sqrt{t}}{2t - t^{2}} + \lim_{t \rightarrow \infty}\frac{t^{2}}{2t - t^{2}}\] Now, the first part will tend to 0 as t approaches infinity, so let's ignore it. The second part will tend to -1 as t approaches infinity, so we have 0 + (-1) = -1. I trust that helps.

OpenStudy (anonymous):

can you explain "Now, the first part will tend to 0 as t approaches infinity, so let's ignore it. The second part will tend to -1 as t approaches infinity, so we have 0 + (-1) = -1." a little more clearly... or step by step :/

OpenStudy (anonymous):

You don't need to do that, you can just take t^2 out of the upper part as well as the second part and take the limit like that:\[\lim_{t \rightarrow \infty} \frac{ t^2(\sqrt{\frac{ 1 }{ t }}+1) }{ t^2(\frac{ 2 }{ t }-1) }\]t^2 cancel, and both fractions tend to 0, giving -1

OpenStudy (anonymous):

This portion is left as an exercise for the reader. No, just kidding. :) I simplified the problem by splitting the fraction into a sum of fractions. Then in solving limits that approach infinity, only the highest degree term (highest power of t) matters, because it will reach infinity before any of the other terms. In the first fraction, the highest degree of t is 2 and it is in the denominator. So as t increases, the denominator continues to get larger than the numerator, and the value of the fraction shrinks, eventually to 0. \[lim_{t \rightarrow \infty}\frac{\sqrt{t}}{2t-t^{2}}\] In the second fraction, the highest degree of t is 2 and it is in both the numerator and the denominator. So as t increases, the numerator and the denominator become approximately the same (except that one is negative and one is positive), and the value of the fraction becomes ever closer to -1. \[lim_{t \rightarrow \infty}\frac{t^2}{2t-t^{2}}\] The sum of these two fractions (0 and -1) is -1.

OpenStudy (anonymous):

holy cow! Okay. I get it now lol :) BTW, thanks so much for 'actually' explaining it :)

OpenStudy (anonymous):

and thanks to you to Ivan! :) Now, I know two ways to solve it. Even better!

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