Continuity at a point

**zhupolongjoe** · October 23rd 2009, 09:08 AM

Set f: (0,1) --> R by:

f(x) = (1/sqrt(x))-sqrt((x+1)/(x))

Can one definite f(0) to make f continuous at ? Explain.

I believe you can set the limit as x approaches 0 equal to f(0) in order to make it continuous.

That is

the limit as f(x) as x->0 is 0, so you could set f(0)=0 to make this happen?

Is this correct?

**tonio** · October 23rd 2009, 09:15 AM

Originally Posted by zhupolongjoe

Set f: (0,1) --> R by:

f(x) = (1/sqrt(x))-sqrt((x+1)/(x))

Can one definite f(0) to make f continuous at ? Explain.

I believe you can set the limit as x approaches 0 equal to f(0) in order to make it continuous.

That is

the limit as f(x) as x->0 is 0, so you could set f(0)=0 to make this happen?

Is this correct?

Yes.

Tonio

**chisigma** · October 23rd 2009, 09:25 PM

With simple passages is...

$f(x)= \frac{1-\sqrt{1+x}}{\sqrt{x}} = \frac{\sqrt{x}}{1+\sqrt{1+x}}$ (1)

All right!... but necessary condition for $f(x)$ to be continous in $x=0$ is...

$\lim_{x \rightarrow 0+} f(x) = \lim_{x \rightarrow 0-} f(x) = f(0)$ (2)

Is that satisfied for (1)?

...

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · October 24th 2009, 03:01 AM

Originally Posted by chisigma

With simple passages is...

$f(x)= \frac{1-\sqrt{1+x}}{\sqrt{x}} = \frac{\sqrt{x}}{1+\sqrt{1+x}}$ (1)

Actually, it is $-\frac{\sqrt{x}}{1+ \sqrt{1+x}}$

All right!... but necessary condition for $f(x)$ to be continous in $x=0$ is...

$\lim_{x \rightarrow 0+} f(x) = \lim_{x \rightarrow 0-} f(x) = f(0)$ (2)

Is that satisfied for (1)?

...

Kind regards

$\chi$ $\sigma$

It is "continuous from the right" if you set f(0)= 0.

And many text books use the term "continuous" to mean that if $\{x_n\}$ is a sequence of points in the domain of f converging to a, then $\lim_{n\to\infty} f(x_n)= f(x)$ .

**chisigma** · October 24th 2009, 08:44 AM

Originally Posted by HallsofIvy

Actually, it is $-\frac{\sqrt{x}}{1+ \sqrt{1+x}}$

In the complex z plain the function...

$f(z)= \frac{\sqrt{z}}{1+\sqrt{1+z}}$ (1)

... is a multivalued function and in particular it has four independent branches connecting two branch points: $z=0$ and $z=-1$ . When $z=x$ , $x$ real the four branches can be represented as...

$f(x)= \frac{\pm \sqrt{x}}{1 \pm \sqrt{1+x}}$ (2)

... and I reported for semplicity sake one of them...

Originally Posted by HallsofIvy

... it is 'continuous from the right' if you set f(0)= 0...

... many text books use the term 'continuous' to mean that if $\{x_n\}$ is a sequence of points in the domain of f converging to a, then $\lim_{n\to\infty} f(x_n)= f(x)$ .

... You say 'many text books use' and that means automatically that 'many text books don't' anf that is enough to make your point of view a little questionable...

Kind regards

$\chi$ $\sigma$

**chisigma** · October 24th 2009, 09:05 AM

A more interesting consideration regarding the multivalued function...

$f(x) = \frac{\pm \sqrt{x}}{1 \pm \sqrt{1+x}}$ (1)

... is that if we choose the sign '-' in denominator is...

$\lim_{x \rightarrow 0+} f(x) = \pm \infty \ne 0$ (2)

... fancy that!

...

Kind regards

$\chi$ $\sigma$

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