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Showing Continuity of a function

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1. Homework Statement
Show that f(x) = x/(1+x^2) is continuous on R


2. Homework Equations
f is continuous at a if for any epsilon > 0, there exists a number delta > 0 such that if |x-a|<delta, then |f(x)-f(a)|<epsilon.


3. The Attempt at a Solution
|f(x) - f(a)| = |x/(1+x^2) - a/(1+a^2)|

= |(x-a)(1-ax)/[(1+x^2)(1+a^2)]|

If I can show that the previous is <= |(x-a)(1-ax)/(1-ax)| then I can cancel the (1-ax) and I'll be done right? I'm not even really sure if I started on this the right way, so any help would be appreciated.
 

Answers and Replies

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one way to try to figure it out...

Let e > 0. Choose d = ....don't know yet...
Suppose |x-a| < d, then
|x/(1 + x^2) - a/(1 + a^2)| = |(x -a + ax(a-x))/((1 + x^2)(1 + a^2))| <= |x-a|/|(1 + x^2)(1 + a^2)| + |ax||x-a|/|(1 + x^2)(1 + a^2)| <= |x-a|/(1 + a^2) + |ax||x-a|/(1 + a^2) <= |x-a| + |x/a||x-a| <= think about what you can say about |x/a|, then try to find a delta that works

this leads to a proof, but there might be an easier way, play with it, goodluck
 
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ok I got it I think. If I set |x-a|<|a|, then |x|-|a|<|a| or |x/a|<2. So the original
|f(x)-f(a)|<3|x-a|. So, if delta = epsilon/3 then the definition given in the first post will be satisfied.

Side note, does that imply uniform continuity since delta depends only on epsilon?
 
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ok I got it I think. If I set |x-a|<|a|, then |x|-|a|<|a| or |x/a|<2. So the original
|f(x)-f(a)|<3|x-a|. So, if delta = epsilon/3 then the definition given in the first post will be satisfied.

Side note, does that imply uniform continuity since delta depends only on epsilon?
good work, but not quite

first, if a = 0, everything is trivial(you should verify), so say a != 0

so we have stuff <= |x-a| + |x/a||x-a|.

as you pointed out, we need |x-a| < |a|. But we also need |x-a| < e/3, so choose
d = min{|a|, e/3}, then if |x-a| < d (this means we have both |x-a| < d <= e/3 and |x-a|<d <= |a| )

stuff <= |x-a| + |x/a||x-a| < e/3 + 2e/3 = e.
 
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Another method, which may be simpler:
If a function f is differentiable at a point x then it must be continuous at x. Thus, it is enough to show that f is differentiable on the whole real line, i.e. that f '(x) exists for all real x.
 

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