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### A Calculus Question

```
Date: 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

```

```
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/
```
Associated Topics:
High School Calculus

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