Course:MATH110/Archive/2010-2011/003/Math Forum/Lecture18

Hi Steffany, So I've worked out your questions.

The limit as X approaches infinity of $sqrt(X^{2}+13x-2)$- x

Think of this as $sqrt((X^{2}+13x-2)-x)/1)$*$((sqrt(X^{2}+13x-2)+x)/(sqrt(X^{2}+13x-2)+x))$

What this step does is it rationalizes the numerator to simplify the equation.

You should get

$(x^{2}+13x-2-x^{2})/(sqrt(X^{2}+13x-2)+x)$

Divide the numerator and denominator by the biggest X power found in the denominator, which in this case is X.

You should get $(X+13-(2/x)-x)/(1+(sqrt13x/x)-(sqrt2/x)+1)$

which simplifies to $13/2$

It's the same types of steps for the second problem, try that,and if it still doesn't work let me know! --BernadetteHii 18:46, 23 October 2010 (UTC)

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Thanks, Bernadette! That makes much more sense now. -- 17:38, 24 October 2010 (UTC)

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I talked to a couple of people in class today who had trouble with this one from the Lecture 18 set yesterday, so I think it is worth mentioning. And yes, I'm on the math forum on a Friday night, so I don't expect to hear back from anyone right away...

The last 2 parts of question 5:

The limit as X approaches infinity of $sqrt(X^{2}+13x-2)$- x and $sqrt(X^{2}+13x-2)$ + x

Any hints to solve these ones would be helpful. I tried several different ways, like pulling out x^2 first, but every time I solve it I get negative infinity for the first one. I haven't had problems with the rest of the ones involving square roots when solving them this way so I'm stuck.

Victoria - about 4(a) - although you may have asked around today and figured this out already - Look at what you are left with after you have pulled the x out of numerator and the x^2 out of the denominator. My hint is to remember that when you square a number x and then take the root of it you are left with... -- 01:04, 23 October 2010 (UTC)

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Has anyone got any suggestions for #4a)? I've come up with so many different answers and none of them are right. Any helpful hints out there? (and is anyone else totally frustrated and feeling completely lost on these?) ANY advice on any of the asymptotes is helpful. I felt like the Wiki notes were helpful for the concept but not for the computations. Thanks- Victoria

Hi Victoria 4a) is similar to the previous questions, so what you have to do first is to sqare-root the denominator then divide both the numerator and the denominator by the largest t power found in the denominator.

My #4a) on the webwork was: The limit as t approaches infinity $(-1t-9)/(sqrt(t^{2}+6t+5)$)

what I did first was I simplify the denominator $(sqrt(t^{2}+6t+5)$) and I got $(t+sqrt(6t)+sqrt(5))$

after that I divide both the denominator $(t+sqrt(6t)+sqrt(5))$ and the numerator $(-1t-9)$ by the largest t power, in this case it would be $(t^{1})$ you will then get $((-1t-9)/t)/(t+(sqrt(6t)/t)+(sqrt(5)/t))$ but since t cancels out we now left with $(-1-(9/t))/(1+(sqrt(6t)/t)+(sqrt(5)/t))$ therefore as the limit t approaches infinity we should get -1.

Let me know if you need a more detail explanation on this question --Narissarasu 17:32, 23 October 2010 (UTC)

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