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Topic: Probability of success calculation
Replies: 17   Last Post: Jun 2, 2011 2:49 PM

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Brian M. Scott

Posts: 1,286
Registered: 12/6/04
Re: As x approaches 5
Posted: Jun 1, 2011 1:55 PM
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On Wed, 01 Jun 2011 10:45:17 -0700, Alexandros Bantis
<ambantis@gmail.com> wrote in
<news:94nc5dFituU1@mid.individual.net> in
alt.math.undergrad:

> I'm working on Salas (10th Ed) Calculus: One and Several Variables.
> Problem #41


> "Define the function at 5 so that it becomes continuous at 5:

> f(x) = sqrt(x+4)-3 / (x-5)

I'm assuming that you mean [sqrt(x + 4) - 3] / (x - 5).

> I tried multiplying by (sqrt(x+4)+3)/(sqrt(x+4)+3), which
> resulted in 1/(sqrt(x+4)-3)


Algebra error: that should be 1 / [sqrt(x + 4) + 3], so that
the limit of f(x) as x approaches 5 is 1/6. To make f
continuous at x = 5, therefore, its definition must be
extended to x = 5 by setting f(5) = 1/6.

> which is still not continuous as x approaches 5.

Incorrect terminology: both what you had and the correct
expression *are* continuous as x approaches 5. The
difference is that one of them has a value at x = 5, and the
other hasn't.

Brian



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