• Apr 29th 2009, 09:11 AM
C.E

1.a. A function f, from the reals to the reals is continuous at 0 with f(0)=0 and another function g from the reals to the reals is bounded then the product fg is continuous at 0.

I think this may be true but don't know how to start a proof, any ideas?

b. If for f from [-1,1] to the reals, sin(f(x)) is continuous (over the reals) then f must be continuous.

I think f(x)=1/x (restricted to [-1,1]) may be a counterexample to this but one problem is it is not defined at 0 which is in [-1,1], is it ok to use it anyway.

I think this may be true but don't know how to start a proof, any ideas?

c. If for f from [-1,1] to the reals f(sinx) is continuous on R then f is continuous on R

2. Find without proof the following limits.
a. cosx/sinx -1/x as x tends to 0 from below.
b. x^x as x tends to 0 from above.

I don't see how to find these limits (short of plugging small numbers into a calculator and seing what seems to be happening). As I won't have a calculator in my exam I was wondering, is there another way?
• Apr 30th 2009, 05:18 AM
Showcase_22
Quote:

b. If for f from [-1,1] to the reals, sin(f(x)) is continuous (over the reals) then f must be continuous.

I think f(x)=1/x (restricted to [-1,1]) may be a counterexample to this but one problem is it is not defined at 0 which is in [-1,1], is it ok to use it anyway.
$\sin \left( \frac{1}{x} \right)$ is discontinuous at the origin since it fluctuates infinitely many times between 1 and -1.

I think the statement is true because of algebra of continuity.
• Apr 30th 2009, 07:42 AM
albi
Quote:

Originally Posted by Showcase_22
$\sin \left( \frac{1}{x} \right)$ is discontinuous at the origin since it fluctuates infinitely many times between 1 and -1.

I think the statement is true because of algebra of continuity.

The function does not exist at 0, so we cannot say if it is continuous or discontinuous...

I also think that the statement is true, but it is not only by continuity...

For example...

$f(x) = \left\{ \begin{array}{c} \frac{1}{|x|} \quad x \neq 0 \\
\pi \quad x = 0 \end{array} \right.$

and $g(x) = sinc(x)$

We see that $\lim_{x \rightarrow 0} g(f(x)) = 0$ and $g(f(0)) = 0$ hence $g \circ f$ is continuous.
• Apr 30th 2009, 09:51 AM
Showcase_22
Quote:

The function does not exist at 0, so we cannot say if it is continuous or discontinuous...
Surely we can say a function is dicontinuous at a point even if it is not defined at a point!

What about $f(x)=\frac{1}{x}$. That's discontinuous at 0 even though it is not defined when x=0.
• Apr 30th 2009, 12:42 PM
albi
Quote:

Originally Posted by Showcase_22
Surely we can say a function is dicontinuous at a point even if it is not defined at a point!

What about $f(x)=\frac{1}{x}$. That's discontinuous at 0 even though it is not defined when x=0.

I know, some authors (for example G.M. Fichtenholtz) say 1/x (for instance) is discontinuous in 0.

Personally I don't like this point of view.
• Apr 30th 2009, 12:46 PM
Showcase_22
What's the consensus?

My lecturer teaches the point of view of Fichtenholtz so that's how i've always done it.
• Apr 30th 2009, 01:42 PM
albi
Quote:

Originally Posted by Showcase_22
What's the consensus?

My lecturer teaches the point of view of Fichtenholtz so that's how i've always done it.

From the point of view of task it doesn't matter...

But from the point of view of topology 1/x is continuous (as a function defined in $\mathbb{R} \setminus \{0\}$)
• Apr 30th 2009, 01:53 PM
Showcase_22
Back to the questions at hand!

$\lim_{x \rightarrow 0-} \left(\frac{\cos x}{\sin x}- \frac{1}{x} \right)$

When x is small $\sin x$ is approximately equal to $x$.

Hence the question becomes:

$\lim_{ x \rightarrow 0-} \left( \frac{ \cos x -1}{x} \right)$

... and do L' Hopital's on that.
• Apr 30th 2009, 03:57 PM
C.E
How would you prove that part b is true?
• Apr 30th 2009, 03:59 PM
putnam120
I don't think anyone has addressed the first question yet so I think I will take that honor.

Let $\epsilon>0$, then there exist a $\delta$ such that $|x|<\delta\Rightarrow |f(x)-f(0)|<\epsilon$. This is just from the definition of continuity at 0. Now since $g$ is bounded let $M=\sup_{x\in\mathbb{R}} |g(x)|$.

From here we have that $|f(x)g(x)-f(0)g(0)|\le|Mf(x)-Mf(0)|=M|f(x)-f(0)|$.

At this point you are basically done (though if this were an exam you would want to actually finish the proof).
• May 1st 2009, 01:55 AM
Showcase_22
Quote:

$\lim_{ x \rightarrow 0+} x^x$
$\lim_{x \rightarrow 0+} e^{ln(x^x)}$

$\lim_{x \rightarrow 0+} e^{x \ ln(x)}$

$e^{\lim_{x \rightarrow 0+} x \ln(x)}$

$e^{\lim_{x \rightarrow 0+} \frac{ln(x)}{\frac{1}{x}}}$

Now it's time for L'Hopital's!

$e^{\lim_{x \rightarrow 0+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}}$

$e^{\lim_{ x \rightarrow 0+} -x}=e^0=1$