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Forum URL: http://www.cut-the-knot.org/cgi-bin/dcforum/forumctk.cgi
Forum Name: College math
Topic ID: 71
#0, Evaluating a limit as x -%3E infinity
Posted by Dave (Guest) on Mar-09-01 at 00:55 AM
I just had a test on this and this was the one problem I couldn't get:

lim as x -> infinity of (sqrt(x - 6) - sqrt(x))


I answered 0 but I wasn't really sure. I thought it may have been 6 or maybe infinity. The usual method I use is dividing everything by x raised to the highest power in the denominator then simplifying. But since the denominator is x^0 this didn't work...


#1, RE: Evaluating a limit as x -%3E infinity
Posted by alexb on Mar-09-01 at 01:03 AM
In response to message #0
LAST EDITED ON Mar-09-01 AT 01:05 AM (EST)

>I just had a test on
>this and this was the
>one problem I couldn't get:
>
>lim as x -> infinity of (sqrt(x - 6) - sqrt(x))
>
>The usual method
>I use is dividing everything
>by x raised to the
>highest power in the denominator
>then simplifying.

You do not have denominator here. But honestly I am bewildered by the implication that there may be a usual method for computing limits. In the very least there is a couple dozens usual methods. You just have to learn them all. No single method will serve you in all circumstances.

>But since
>the denominator is x^0 this
>didn't work...

Right. I may give two advices:

1. Get into the habit of consulting your textbook.
2. Divide and multiply your expression by the sum (sqrt(x - 6) + sqrt(x))