Continuous functions: f(x) = x^2 / (x^2 - 1) on (1, infty)

[Jenny4]

Hiya guys, please help me on this question. I've wasted hours on this and I haven't got anywhere!

Show that the rational function $\displaystyle f: (1, \infty) \to \Re$ defined by

$\displaystyle f(x) = \frac{x^2}{x^2 - 1}$, $\displaystyle x \in (1, \infty)$,

is continuous on the interval $\displaystyle (1, \infty)$


Thank you,
Jenny
x

 

[galactus]

Re: Continuous functions

The rational function is not continuous at x = -1 and x = 1. That is rather obvious.

Remember, a rational function is continuous everywhere except where the denominator is 0.

In this case, at x = -1 and x = 1.

Here is a proof that f/g is continuous at c if g(c) does not equal 0 and is discontinuous at c if g(c)=0.

If g(c)=0, then f/g is discontinuous at c because f(c)/g(c) is undefined. Assume that g(c) does not equal 0.

Then we have to show that $\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=\frac{f(c)}{g(c)}$

Since f and g are continuous at c, then by limit properties:

$\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=\frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}g(x)}=\frac{f(c)}{g(c)}$.

Does that help any?. You can always try the limit as $\displaystyle x\to{\infty}$

 

[Jenny4]

The rational function is not continuous at x = -1 and x = 1. That is rather obvious.

Remember, a rational function is continuous everywhere except where the denominator is 0.

In this case, at x = -1 and x = 1.

Here is a proof that f/g is continuous at c if g(c) does not equal 0 and is discontinuous at c if g(c)=0.

If g(c)=0, then f/g is discontinuous at c because f(c)/g(c) is undefined. Assume that g(c) does not equal 0.

Then we have to show that $\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=\frac{f(c)}{g(c)}$

Since f and g are continuous at c, then by limit properties:

$\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=\frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}g(x)}=\frac{f(c)}{g(c)}$.

Does that help any?. You can always try the limit as $\displaystyle x\to{\infty}$

Hi. Thanks for the reply.

Unfortunately, I still don't understand what I'm supposed to do.

If we show that $\displaystyle \lim_{x \to c} \left( \frac{x^2}{x^2 - 1} \right) = f(c)$ for any arbitrary c, then does that all we need to prove that the function is continuous on the interval? I thought it required something more and this would only be part of an 'if and only if' proof, but evidently, my knowledge on the subject is pretty poor.

Anyway, if this is all we need to prove, I don't know how I could go about proving that $\displaystyle \lim_{x \to c} x^2 = f(c)$ and $\displaystyle \lim_{x \to c} \frac{1}{x^2-1} = f(c)$ then by if you multiply both functions together, that should also be continuous at f(c), which is what we want, right?

Please help! Thank you x

 

[pka]

If you will just take a moment and go to your calculus textbook, it can be anyone, and then look in the section on continuity. You will find there a theorem that states that if each of f & g is continuous at x=c and \(\displaystyle \[g(c) \not= 0\) then $\displaystyle \frac {f} {g}$ is also continuous at x=c. Note that is a point-wise condition.

Both functions $\displaystyle x^2 \,\& \,1 - x^2$ are continuous at each point of the interval $\displaystyle (1,\infty )$, the second is not zero there, therefore by the theorem their quotient is continuous at each point of the interval $\displaystyle (1,\infty )$.

That works this problem. Now what don’t you understand?

 

[Jenny4]

If you will just take a moment and go to your calculus textbook, it can be anyone, and then look in the section on continuity. You will find there a theorem that states that if each of f & g is continuous at x=c and \(\displaystyle \[g(c) \not= 0\) then $\displaystyle \frac {f} {g}$ is also continuous at x=c. Note that is a point-wise condition.

Both functions $\displaystyle x^2 \,\& \,1 - x^2$ are continuous at each point of the interval $\displaystyle (1,\infty )$, the second is not zero there, therefore by the theorem their quotient is continuous at each point of the interval $\displaystyle (1,\infty )$.

That works this problem. Now what don’t you understand?

Thank you for your reply.

I don't understand what I'm supposed to do to show that $\displaystyle \lim_{x \to c} x^2 = f(c)$? I've tried using the delta-epsilon definition, but didn't get anywhere.

I know I'm being a pain, but I've spent so much time on this question, and it's really bugging me.

Any further pointers will be much appreciated.

Jenny x

 

[galactus]

It appears you're making it harder than need be. Look at the previous posts and don't worry about an epsilon-delta proof.

It isn't that involved. Just use the theorems.

 

[pka]

If you insist upon using the delta-epsilon definition, then find yourself a proof of the theorem that I quoted above. Just follow what is done there and make it particular.

But I really see no reason to reinvent the proof.