• October 17th 2009, 07:22 PM
450081592
Prove that if there is a number B such that |f(x)| <= B for all x not = 0, then lim x approaches 0 xf(x) = 0. Note, exercise 43-46 are special cases of this gerenal result.

Special cases are:
43. lim x approaches 0 sin (1/x) = 0;
44. lim x approaches pi (x - pi) cos^2 [1/(x-pi)] = 0;
45. lim x approaches 1 |x -1| sinx = 0;
46.
f(x) = {1 if x rational
f(x) = {0, if x irrational then lim x approaches 0 xf(x) = 0.

This is the hardest question in my assgnment, no one in my study group is able to do it, I really need anyone who can give me some hints on this, thanks!
• October 17th 2009, 08:42 PM
tonio
Quote:

Originally Posted by 450081592
Prove that if there is a number B such that |f(x)| <= B for all x not = 0, then lim x approaches 0 xf(x) = 0. Note, exercise 43-46 are special cases of this gerenal result.

Special cases are:
43. lim x approaches 0 sin (1/x) = 0;
44. lim x approaches pi (x - pi) cos^2 [1/(x-pi)] = 0;
45. lim x approaches 1 |x -1| sinx = 0;
46.
f(x) = {1 if x rational
f(x) = {0, if x irrational then lim x approaches 0 xf(x) = 0.

This is the hardest question in my assgnment, no one in my study group is able to do it, I really need anyone who can give me some hints on this, thanks!

0 <= |xf(x)| = |x||f(x)| <= |x|B --> 0 when x --> 0, by arithmetic of limits so the squeezing (or the sandwich) lemma gives you the result.

Tonio
• October 17th 2009, 08:51 PM
Prove It
Quote:

Originally Posted by 450081592
Prove that if there is a number B such that |f(x)| <= B for all x not = 0, then lim x approaches 0 xf(x) = 0. Note, exercise 43-46 are special cases of this gerenal result.

Special cases are:
43. lim x approaches 0 sin (1/x) = 0;
44. lim x approaches pi (x - pi) cos^2 [1/(x-pi)] = 0;
45. lim x approaches 1 |x -1| sinx = 0;
46.
f(x) = {1 if x rational
f(x) = {0, if x irrational then lim x approaches 0 xf(x) = 0.

This is the hardest question in my assgnment, no one in my study group is able to do it, I really need anyone who can give me some hints on this, thanks!

45.
$0 \leq \sin{x} \leq 1$

$0|x - 1| \leq |x - 1|\sin{x} \leq 1|x - 1|$

$0 \leq |x - 1|\sin{x} \leq |x - 1|$

Now take the limit as $x \to 1$.
• October 17th 2009, 09:04 PM
450081592
Quote:

Originally Posted by tonio
0 <= |xf(x)| = |x||f(x)| <= |x|B --> 0 when x --> 0, by arithmetic of limits so the squeezing (or the sandwich) lemma gives you the result.

Tonio

could you tell me more about it, cause we didn;t learn the lemma therom yet, how do I apply it to this proof?
• October 17th 2009, 09:05 PM
450081592
Quote:

Originally Posted by Prove It
45.
$0 \leq \sin{x} \leq 1$

$0|x - 1| \leq |x - 1|\sin{x} \leq 1|x - 1|$

$0 \leq |x - 1|\sin{x} \leq |x - 1|$

Now take the limit as $x \to 1$.

43 to 46 is not the questions, they are the result that are already proven, but thanks though
• October 18th 2009, 12:54 AM
tonio
Quote:

Originally Posted by 450081592
could you tell me more about it, cause we didn;t learn the lemma therom yet, how do I apply it to this proof?

The sandwich lemma: let f,g,h be three functions defined in some neighborhood of a point w (excetp, perhaps, in w itself) and s.t. g(x) <= f(x) <= h(x) for all x in this neighborhood. If g(x) --> L and h(x) --> L as x --> w, then also f(x) --> L as x --> w.

There's a corresponding very wll known lemma in limits of sequences, too.

Tonio