Is My Epsilon-Delta Proof of a Limit Correct?

[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 :

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

<br /> \forall\epsilon&gt;0\ \exists\ \delta&gt;0\ such\ that\ 0&lt;|x - 10| &lt; \delta \implies |x^2 - 100|&lt; \epsilon<br />
I observe the following
|x-10| = |x + (-10)| and by triangle inequality, |x|+|-10| &gt; |x-10| We will also note that |x| + |-10| = |x| + |10|\implies|x|+|10|&gt;|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||&gt;|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2\implies\epsilon &gt; |x|^2 - |10|^2 Using this we can see that,\frac{\epsilon+|10|^2}{|x|}&gt;|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}&gt;|10| \ and\ since\ we\ know:\ |x|+|10| &gt; |x-10|,
\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} &gt; |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|}

 

[STEMucator]

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 :

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

<br /> \forall\epsilon&gt;0\ \exists\ \delta&gt;0\ such\ that\ 0&lt;|x - 10| &lt; \delta \implies |x^2 - 100|&lt; \epsilon<br />
I observe the following
|x-10| = |x + (-10)| and by triangle inequality, |x|+|-10| &gt; |x-10| We will also note that |x| + |-10| = |x| + |10|\implies|x|+|10|&gt;|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||&gt;|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2\implies\epsilon &gt; |x|^2 - |10|^2 Using this we can see that,\frac{\epsilon+|10|^2}{|x|}&gt;|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}&gt;|10| \ and\ since\ we\ know:\ |x|+|10| &gt; |x-10|,
\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} &gt; |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|}

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?

 

[DarthRoni]

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

 

[STEMucator]

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

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

Now what can you conclude?

$δ|x+10| < δ(δ+20)$

 

[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

 

[STEMucator]

Oh is see ! Suppose |x-10|&lt;1\implies |x+10| &lt; 1 + 2|10|\ and\ |x^2 - 100| &lt; \delta(\delta + 20) So we can take \epsilon = \delta^2 + 20\delta \implies \epsilon = (1+2|10|)^2 + 20(1+2|10|) \implies \frac{\epsilon}{1+2|10|} - 20 = \delta

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 $δ = min\{1, \frac{ε}{21}\}$

 

[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 ?

 

[STEMucator]

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 ?

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

 

[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