A Calculus QuestionDate: 02/09/98 at 18:52:51 From: Scott Nash Subject: A Calculus question Could you help me solve this question? I figured it out to work out to -1/2. lim (x - (x^2 + x)^(1/2)) x --> infinity Thank you for your assistance. Date: 02/09/98 at 21:34:22 From: Doctor Sam Subject: Re: A Calculus question Scott, You are correct. Here is a method. You cannot find this limit either by substitution (since infinity isn't a real number) or by finding the limit of each term separately. Somehow we need to combine these two terms into a single expression. The presence of the square root might remind you of problems with fractions with expressions like this one in the denominator. In those problems you had to rationalize the fraction in order to get rid of the denominator. In this case, we will use exactly the same technique but with the goal of getting the unrationalized fraction back: Start with (x - sqrt(x^2+x) ) and multiply it by the fraction (x + sqrt(x^2+x)) --------------------------- (x + sqrt(x^2+x)) This gives the new fraction: (x^2 - (x^2+x)) -x -------------------- which simplifies to ----------------- (x + sqrt(x^2+x)) (x + sqrt(x^2+x)) While this is bad for working with fractions, it is great for working with limits! We still can't just take the limit of numerator and denominator but we can use one more Calculus trick: divide both the numerator and the denominator by x: (-x/x) -1 ----------------- = ------------------ by distributing the [ x + sqrt(x^2+x)]/x 1 + [sqrt(x^2+x)/x] division Finally, you can bring the x inside the square root since 1/x = sqrt (1/x^2): -1 = -------------- 1 + sqrt(1 + 1/x) Finally we have something that will work! Now take the limit as x grows to infinity. The 1's don't care about x...only 1/x is affected and it approaches zero. This leaves the fraction -1 / [1 + sqrt(1)] = -1/2 Well done, Scott. -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/