2 Replies - 1227 Views - Last Post: 09 September 2013 - 04:52 PM

#1 TheYoungWolf  Icon User is offline

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Determine Infinite limits in calculus

Posted 28 August 2013 - 05:28 PM

Hey guys I'm doing ok in calculus but am still totally stumped when it comes to finding the infinite limit of a finite number. My textbook does not explain it well, and whenever I search for this topic on the internet I usually just get problems where the limit is infinity (which I find much easier to solve). All my efforts have been to draw a number line, and try to visualize the problem out. Some examples are below, and I would really appreciate it if someone could work them and explain to me how they came to negative infinity or positive infinity.

Determine if the infinite limit of the equation is positive or negative....the limit, as x approaches -4 from the left, (x + 5)/(x-4).

Determine if the infinite limit of the equation is positive or negative....the limit, as x approaches 2 from the right, (x^2 - 2x - 8)/(x^2 - 5x + 6)

Thanks!

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Replies To: Determine Infinite limits in calculus

#2 mojo666  Icon User is offline

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Re: Determine Infinite limits in calculus

Posted 29 August 2013 - 10:13 AM

Quote

My textbook does not explain it well, and whenever I search for this topic on the internet I usually just get problems where the limit is infinity (which I find much easier to solve).


Ummm... for the two examples you give, the limit is infinity. It even tells you at the beginning of the problems. It is just asking you whether the limit is positive or negative at the given value.

For the first one, when x is 4, the numerator is 9 and the denominator is 0. Slightly more than 4, it is numerator>9 and denominator>0. Slightly less than 4, it is numerator<9 and denominator<0. You should be able to figure out the sign from this information.
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#3 jjl  Icon User is offline

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Re: Determine Infinite limits in calculus

Posted 09 September 2013 - 04:52 PM

You can apply L'hopitals rule to the second, since direct plug -n-chug will be inconclusive.

i.e.

Take the derivitive of the numerator and the denominator

n(x) / d(x) = (x^2 - 2x - 8) / (x^2 - 5x + 6)
n'(x) / d'(x) = (2x - 2) / (2x - 5)

Now do a plug n' chug with x = 2.

n'(2) / d'(2) = (4 - 2) / (4 - 5) = 2 / -1 = -2


More on L'hopitals rule (http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx)


scratch that, I can't do basic arithmetic. You can plug and chug directly, you will find that the numerator != 0, and the deninator = infinity, impling the limit is either positive or negitive infinity

This post has been edited by jjl: 09 September 2013 - 04:55 PM

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