Show by means of example that lim/x→a/[f(x)g(x)] may exist

  • Thread starter Thread starter MathewsMD
  • Start date Start date
  • Tags Tags
    Example Means
MathewsMD
Messages
430
Reaction score
7
Show by means of example that lim/x→a/[f(x)g(x)] may exist even though neither lim/x→a/f(x) nor lim/x→a/g(x) exists.

I have tried using examples such as piecewise functions and rational functions, but can never validate the statement.

Any guidance and help would be great.

Thanks.
 
Physics news on Phys.org
Let f(x) be any crazy function. Let g(x) = 1/f(x).
 
mathman said:
Let f(x) be any crazy function. Let g(x) = 1/f(x).

You forgot to insert Arildno's corollary:
"Let f(x) be any crazy function. Let g(x) = 1/f(x). THEN, g(x) is most likely also a crazy function"

Not very useful in this context, of course, but the result is beautiful, nonetheless. :smile:
 
MathewsMD said:
Show by means of example that lim/x→a/[f(x)g(x)] may exist even though neither lim/x→a/f(x) nor lim/x→a/g(x) exists.

I have tried using examples such as piecewise functions and rational functions, but can never validate the statement.

Any guidance and help would be great.

Thanks.

Consider the function which is equal to 1 if its argument is rational and 0 otherwise.
 
pasmith said:
Consider the function which is equal to 1 if its argument is rational and 0 otherwise.
I presume you mean "let f(x)= 1 if x is rational, 0 if x is irrational.

And then let g(x)= 0 if x is rational, 1 if x is irrational.


fg(x)= 0 for all x so it trivially differentiable.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top