Solving Exercise: Proving Continuity of a Function at 1

[Bueno]

Hello everyone!

I'm having some trouble to solve the following exercise:

Supposing that \(\displaystyle |f(x) - f(1)|≤ (x - 1)^2\) for every \(\displaystyle x \).
Show that \(\displaystyle f\) is continuous at \(\displaystyle 1\)

(Sorry if the text seems a bit weird, but it's because I'm still getting used to translate all these math-related terms to english.)

I know that if f is continuous at 1, the following will be truth:
\(\displaystyle
0<|x-1|< \delta\) \(\displaystyle ⇒\) \(\displaystyle |f(x) - f(1)| < \epsilon\)

I thought of choosing \(\displaystyle \delta = \epsilon/2(x-1)^2\), then I would find that \(\displaystyle |f(x) - f(1)| < \epsilon/2 < \epsilon\)

But, as far as I know, choosing a \(\displaystyle \delta\) that depends on \(\displaystyle x\) is wrong.
I really don't know what to do.

Thank you,

Bueno.

 

[chisigma]

Hello everyone!

I'm having some trouble to solve the following exercise:

Supposing that \(\displaystyle |f(x) - f(1)|≤ (x - 1)^2\) for every \(\displaystyle x \).
Show that \(\displaystyle f\) is continuous at \(\displaystyle 1\)

(Sorry if the text seems a bit weird, but it's because I'm still getting used to translate all these math-related terms to english.)

I know that if f is continuous at 1, the following will be truth:
\(\displaystyle
0<|x-1|< \delta\) \(\displaystyle ⇒\) \(\displaystyle |f(x) - f(1)| < \epsilon\)

I thought of choosing \(\displaystyle \delta = \epsilon/2(x-1)^2\), then I would find that \(\displaystyle |f(x) - f(1)| < \epsilon/2 < \epsilon\)

But, as far as I know, choosing a \(\displaystyle \delta\) that depends on \(\displaystyle x\) is wrong.
I really don't know what to do.

Thank you,

Bueno.

On the basis of your hypothesis is... $\displaystyle |\frac{f(x)-f(1)}{x-1}| \le |x-1| \implies \lim_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1} = 0$ (1)... so that the derivative $\displaystyle f^{\ '}(x)$ in x=1 exists and is $\displaystyle f^{\ '} (1)=0$. That means that in x=1 f(x) must be continous... Kind regards

$\chi$ $\sigma$

 

[Bueno]

The main problem is I can't use derivative techniques to solve this problem. The professor only talked about limits and their properties in class, and he'd like we figure out how to prove this only using these tools.

Thank you,

Bueno

 

[Fernando Revilla]

The main problem is I can't use derivative techniques to solve this problem. The professor only talked about limits and their properties in class, and he'd like we figure out how to prove this only using these tools.

Thank you,

Bueno

In that case choose $\delta=\sqrt{\epsilon}$: $$|x-1|<\delta\Rightarrow |x-1|^2<\epsilon\Rightarrow \left|f(x)-f(1)\right|<\epsilon$$

 

[Bueno]

In that case choose $\delta=\sqrt{\epsilon}$: $$|x-1|<\delta\Rightarrow |x-1|^2<\epsilon\Rightarrow \left|f(x)-f(1)\right|<\epsilon$$

That seems to work, thank you!

But I have to say I'm a bit confused by this kind of proof.
Choosing an appropriate value for delta seems to do the work, but is there any kind of manipulation or technique I can do to make the value I have do choose become more clear?

Thank you,

Bueno

 

[Fernando Revilla]

but is there any kind of manipulation or technique I can do to make the value I have do choose become more clear?

Unfortunately, there is no fixed set of rules for this kind of problems.

 

[Ackbach]

Unfortunately, there is no fixed set of rules for this kind of problems.

Well, there might be, actually. Check http://www.mathhelpboards.com/f49/method-proving-some-non-linear-limits-4149/ out. I realize it's about proving limits, not continuity. But really, the definitions are so similar that instead of $L$ you could just say $f(1)$ (in your case), and you can change the appropriate inequalities not to be strict, and I think you'd have the same logical structure.

 

[Fernando Revilla]

Well, there might be, actually. Check http://www.mathhelpboards.com/f49/method-proving-some-non-linear-limits-4149/ out.

No problem, we have not yet agreed what we mean by fixed set of rules. By the way, an excellemt post.