Epsilon Delta Proof of a Limit

1. Oct 11, 2013

DarthRoni

Hey there, I'm new to this forum. Today I thought I would brush up on my calculus.
I would just like to know if my method is correct. Is there an easier way to prove this ?
By the way, it's my first time using LaTeX, so bear with me.

I am trying to prove the following :

$$\lim_{x\rightarrow 10} {x^2} = 100$$
So, I must find a δ in which the following holds

$$\forall\epsilon>0\ \exists\ \delta>0\ such\ that\ 0<|x - 10| < \delta \implies |x^2 - 100|< \epsilon$$
I observe the following
$$|x-10| = |x + (-10)|$$ and by triangle inequality, $$|x|+|-10| > |x-10|$$ We will also note that $$|x| + |-10| = |x| + |10|\implies|x|+|10|>|x-10|$$
Now we can find a δ in terms of ε. By reverse triangle inequality,
$$|x^2 - 100|\geq||x^2| - |100||\ and\ ||x^2| - |100||>|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2$$$$\implies\epsilon > |x|^2 - |10|^2$$ Using this we can see that,$$\frac{\epsilon+|10|^2}{|x|}>|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}>|10| \ and\ since\ we\ know:\ |x|+|10| > |x-10|,$$
$$\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} > |x-10|$$
So take $$\delta = \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|}$$

Is this correct? I would be really grateful if I got some feedback!
Can we also say the function is continuous $\forall c\in\mathbb R$ in the following way.
$$\lim_{x\rightarrow c} {x^2} = c^2$$ just take $$\delta = \frac{\epsilon+|c|^2}{|x|} + \frac{\epsilon-|x|^2}{|c|}$$

Last edited: Oct 11, 2013
2. Oct 11, 2013

Zondrina

That seems incredibly overcomplicated.

$|x^2-100| = |x-10||x+10| < δ|x+10|$

Now by the triangle inequality:

$|x+10| = |x - 10 + 20| ≤ |x-10| + 20 < δ + 20$

Now bound delta, what can you conclude?

3. Oct 11, 2013

DarthRoni

Can we just say $\epsilon = \delta|x+10|\implies\frac{\epsilon}{|x+10|}=\delta$

4. Oct 11, 2013

Zondrina

Bound delta by the simplest real number you can think of $δ ≤ 1$.

Now what can you conclude?

$δ|x+10| < δ(δ+20)$

5. Oct 11, 2013

DarthRoni

I see now ! $$take\ \delta ≤ 1\implies |x+10|≤ 1+2|10|$$ and so since we can say$$\epsilon = \delta |x+10| \implies \frac{\epsilon}{1+2|10|} = \delta$$

Last edited: Oct 11, 2013
6. Oct 11, 2013

Zondrina

You could go as far as to say:

$δ(δ+20) ≤ 21δ$ because $δ ≤ 1$.

Then you could conclude:
$21δ ≤ ε$
$δ ≤ \frac{ε}{21}$

So $δ = min\{1, \frac{ε}{21}\}$

7. Oct 11, 2013

DarthRoni

Thanks so your help I realized my answer was wrong after I posted it, Why is it we cannot use x to represent out epsilon ?

8. Oct 11, 2013

Zondrina

$x$ has nothing to do with $ε$. The objective is to get a $δ(ε)$.

9. Oct 11, 2013

DarthRoni

I think I understand now, we are trying to find a bound on $|x-c|$ and it doesn't make sense to include $x$ in the value for $\delta$