|
|
Re: As x approaches 5
Posted:
Jun 1, 2011 1:55 PM
|
|
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
|
|
