Proving Definition of Continuous Function: Hi, Carla!

[Carla1985]

Hi, I have my exam tomorrow and have been doing the practice questions. However we don't get the answers to these questions so I am lost as to whether I am doing them right, also I am stuck at a few points, particularly with the definition questions.

\[

1.)Prove\ lim_{x \to \infty}\frac{2n+3}{n+1}=2 \\
I\ did:\ |a_n-a_m|=|(a_n-a)-(a_m-a)|\leq|a_n-a|+|a_m-a| \\
=|\frac{2n+3}{n+1}-2|+|\frac{2m+3}{m+1}-2|=|\frac{1}{n+1}|+|\frac{1}{m+1}|\leq\frac{1}{n}+\frac{1}{m}=\frac{m+n}{mn}\\
\text{not sure how to get that less than epsilon. usually i have m-n so can take out the m's.}\\
\ \\
\ \\
2.)Prove\ by\ definition\ f(x)=|x|\ is\ continuous\ at\ x_0=0\\
I\ have\ the\ definition\ as\ \forall\epsilon>0\ \exists\delta>0\ such\ that\ \forall y\in\mathbb{R}\ |x_0-y|<\delta\ implies\ |f(x_0)-f(y)|<\epsilon"\\
so\ I\ have:\ |0-y|=|-y|<\delta\ \rightarrow\ ||0|-|y||=|-y|<\epsilon\\
I\ know\ I'm\ supposed\ to\ relate\ delta\ and\ epsilon\ together\ somehow\ but\ cant\ for\ the\ life\ of\ me\ figure\ out\ why.\ The\ notes\ we\ have\ are\ very\ confusing\\

\]

Thank you for the help. I would be so lost without this forum :)
Carla x

 

[chisigma]

Hi, I have my exam tomorrow and have been doing the practice questions. However we don't get the answers to these questions so I am lost as to whether I am doing them right, also I am stuck at a few points, particularly with the definition questions.

\[

1.)Prove\ lim_{x \to \infty}\frac{2n+3}{n+1}=2 \\
I\ did:\ |a_n-a_m|=|(a_n-a)-(a_m-a)|\leq|a_n-a|+|a_m-a| \\
=|\frac{2n+3}{n+1}-2|+|\frac{2m+3}{m+1}-2|=|\frac{1}{n+1}|+|\frac{1}{m+1}|\leq\frac{1}{n}+\frac{1}{m}=\frac{m+n}{mn}\\
\text{not sure how to get that less than epsilon. usually i have m-n so can take out the m's.}\\

Is...

$$\frac{2\ n + 3}{n+1} = \frac{2 + \frac{3}{n}}{1 + \frac{1}{n}}\ (1)$$

Now what does it happen if $n \rightarrow \infty$?... Kind regards$\chi$ $\sigma$

 

[Prove It]

Hi, I have my exam tomorrow and have been doing the practice questions. However we don't get the answers to these questions so I am lost as to whether I am doing them right, also I am stuck at a few points, particularly with the definition questions.

\[

1.)Prove\ lim_{x \to \infty}\frac{2n+3}{n+1}=2 \\
I\ did:\ |a_n-a_m|=|(a_n-a)-(a_m-a)|\leq|a_n-a|+|a_m-a| \\
=|\frac{2n+3}{n+1}-2|+|\frac{2m+3}{m+1}-2|=|\frac{1}{n+1}|+|\frac{1}{m+1}|\leq\frac{1}{n}+\frac{1}{m}=\frac{m+n}{mn}\\
\text{not sure how to get that less than epsilon. usually i have m-n so can take out the m's.}\\
\ \\
\ \\
2.)Prove\ by\ definition\ f(x)=|x|\ is\ continuous\ at\ x_0=0\\
\text{I have the definition as }"\forall\epsilon>0\ \exists\delta>0\ such\ that\ \forall y\in\mathbb{R}\ |x_0-y|<\delta\ implies\ |f(x_0)-f(y)|<\epsilon"\\
so\ I\ have:\ |0-y|=|-y|<\delta\ \rightarrow\ ||0|-|y||=|-y|<\epsilon\\
\text{I know I'm supposed to relate delta and epsilon together somehow but can't for the life of me figure out why. The notes we have are very confusing}\\
\ \\
\ \\
3.) Prove\ using\ the\ sequence\ definition\ that\ f(x)=10x^2\ is\ continuous\ at\ x_0=0\\
I\ have:\ take\ any\ sequence\ x_n\ converging\ to\ 0.\ Then\ f(x_n)=10x_n^2\ converges\ to\ f(x_0)=10*0^2=0\ so\ it\ is\ continuous.\\
\text{is that sufficient?}

\]

Thank you for the help. I would be so lost without this forum :)
Carla x

To prove [math]\displaystyle \begin{align*} \lim_{n \to \infty} \frac{2n+3}{n+1} = 2 \end{align*}[/math] you need to show [math]\displaystyle \begin{align*} n > N \implies \left| \frac{2n+3}{n+1} - 2 \right| < \epsilon \end{align*}[/math], so by working on the second inequality we find

[math]\displaystyle \begin{align*} \left| \frac{2n+3}{n+1} - 2 \right| &< \epsilon \\ \left| \frac{2n + 3 - 2\left( n + 1 \right) }{n + 1} \right| &< \epsilon \\ \left| \frac{2}{n + 1} \right| &< \epsilon \\ \frac{2}{| n + 1|} &< \epsilon \\ 2 &< \epsilon |n + 1| \\ \frac{2}{\epsilon} &< | n + 1| \\ \frac{2}{\epsilon} < | n + 1| &< | n | + |1| \textrm{ by the Triangle Inequality} \\ \frac{2}{\epsilon} &< |n| + 1 \\ \frac{2}{\epsilon} - 1 &< |n| \\ \frac{2}{\epsilon} - 1 &< n \textrm{ since } n \to \infty \implies n > 0 \end{align*}[/math]

So if you let [math]\displaystyle \begin{align*} N = \frac{2}{\epsilon} - 1 \end{align*}[/math] and reverse the process, you will have your proof :)

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2.)Prove\ by\ definition\ f(x)=|x|\ is\ continuous\ at\ x_0=0\\
\text{I have the definition as }"\forall\epsilon>0\ \exists\delta>0\ such\ that\ \forall y\in\mathbb{R}\ |x_0-y|<\delta\ implies\ |f(x_0)-f(y)|<\epsilon"\\
so\ I\ have:\ |0-y|=|-y|<\delta\ \rightarrow\ ||0|-|y||=|-y|<\epsilon\\
\text{I know I'm supposed to relate delta and epsilon together somehow but can't for the life of me figure out why. The notes we have are very confusing

A function is continuous at a point if it is defined at that point and the limit of the function as you approach that point is equal to the function value. It's pretty obvious that the function is defined at [math]\displaystyle \begin{align*} x = 0 \end{align*}[/math] and that [math]\displaystyle \begin{align*} f(0) = |0| = 0 \end{align*}[/math]. So for [math]\displaystyle \begin{align*} f(x) = |x| \end{align*}[/math] to be continuous at [math]\displaystyle \begin{align*} x = 0 \end{align*}[/math] you need to show [math]\displaystyle \begin{align*} |x - 0| < \delta \implies \left| |x| - 0 \right| < \epsilon \end{align*}[/math]. Working on the second equality we have

[math]\displaystyle \begin{align*} \left| |x| - 0 \right| &< \epsilon \\ \left| | x | \right| &< \epsilon \\ |x| &< \epsilon \\ |x - 0 | &< \epsilon \end{align*}[/math]

So that means we can let [math]\displaystyle \begin{align*} \delta = \epsilon \end{align*}[/math] and reverse the process, this will complete the proof :)

 

[Carla1985]

Aaaaah, ofc. I was using the cauchy definition as opposed to the convergence one. Explains where I was stuck. Thank you so much for the help :)

 

[I like Serena]

1) Prove $$\lim_{x \to \infty}\frac{2n+3}{n+1}=2$$

I did: $$|a_n-a_m|=|(a_n-a)-(a_m-a)|\leq|a_n-a|+|a_m-a| \\
=|\frac{2n+3}{n+1}-2|+|\frac{2m+3}{m+1}-2|=|\frac{1}{n+1}|+|\frac{1}{m+1}|\leq\frac{1}{n}+\frac{1}{m}=\frac{m+n}{mn}$$

not sure how to get that less than epsilon. usually i have m-n so can take out the m's.

Suppose m, n > N, then
$$\frac{1}{n}+\frac{1}{m} < \frac 1 N + \frac 1 N = \frac 2 N$$

So pick $$N = \frac 2 \varepsilon$$.

 

[solakis1]

So if you let [math]\displaystyle \begin{align*} N = \frac{2}{\epsilon} - 1 \end{align*}[/math] :)

This is wrong because only some values of ε will produce natural Nos N

For example,for :[math]\epsilon=\frac{3}{5}[/math] we get : [math] N=\frac{7}{3}[/math].

For :[math]\epsilon=\frac{5}{7}[/math] we get :[math] N = \frac{9}{5}[/math]

While for : [math]\epsilon=3[/math] we get:[math] N= -\frac{1}{3}[/math]

Only for ε=1 we get : N=1.

But we wnt to be able to find an N for all values of ε

 

[solakis1]

Suppose m, n > N, then
$$\frac{1}{n}+\frac{1}{m} < \frac 1 N + \frac 1 N = \frac 2 N$$

So pick $$N = \frac 2 \varepsilon$$.

Again this is wrong not all values of ε will produce natural Nos N

 

[Evgeny.Makarov]

Suppose m, n > N, then
$$\frac{1}{n}+\frac{1}{m} < \frac 1 N + \frac 1 N = \frac 2 N$$

So pick $$N = \frac 2 \varepsilon$$.

Again this is wrong not all values of ε will produce natural Nos N

Defining $$N = \frac 2 \varepsilon$$ is a little sloppy but harmless. It is clear than one may take any natural number $\ge2/\varepsilon$, e.g., $N=\lceil 2/\varepsilon\rceil$. Then $N\ge2/\varepsilon$, so $1/n+1/m<2/N\le\varepsilon$.

 

[solakis1]

Defining $$N = \frac 2 \varepsilon$$ is a little sloppy but harmless. It is clear than one may take any natural number $\ge2/\varepsilon$, e.g., $N=\lceil 2/\varepsilon\rceil$. Then $N\ge2/\varepsilon$, so $1/n+1/m<2/N\le\varepsilon$.

Not sloppy at all and very harmfulll if one does not know on what bases one should be able, given any real No, to choose a natural No N.

Besides that, he does not prove that lima_n=2 ,but he tries to prove unsuccessfully that :

if lim a_n=2 then the sequence is a Cauchy sequence

 

[solakis1]

To prove [math]\displaystyle \begin{align*} \lim_{n \to \infty} \frac{2n+3}{n+1} = 2 \end{align*}[/math] you need to show [math]\displaystyle \begin{align*} n > N \implies \left| \frac{2n+3}{n+1} - 2 \right| < \epsilon \end{align*}[/math], so by working on the second inequality we find

[math]\displaystyle \begin{align*} \left| \frac{2n+3}{n+1} - 2 \right| &< \epsilon \\ \left| \frac{2n + 3 - 2\left( n + 1 \right) }{n + 1} \right| &< \epsilon \\ \left| \frac{2}{n + 1} \right| &< \epsilon \\ \frac{2}{| n + 1|} &< \epsilon \\ 2 &< \epsilon |n + 1| \\ \frac{2}{\epsilon} &< | n + 1| \\ \frac{2}{\epsilon} < | n + 1| &< | n | + |1| \textrm{ by the Triangle Inequality} \\ \frac{2}{\epsilon} &< |n| + 1 \\ \frac{2}{\epsilon} - 1 &< |n| \\ \frac{2}{\epsilon} - 1 &< n \textrm{ since } n \to \infty \implies n > 0 \end{align*}[/math]

So if you let [math]\displaystyle \begin{align*} N = \frac{2}{\epsilon} - 1 \end{align*}[/math] and reverse the process, you will have your proof :)

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A function is continuous at a point if it is defined at that point and the limit of the function as you approach that point is equal to the function value. It's pretty obvious that the function is defined at [math]\displaystyle \begin{align*} x = 0 \end{align*}[/math] and that [math]\displaystyle \begin{align*} f(0) = |0| = 0 \end{align*}[/math]. So for [math]\displaystyle \begin{align*} f(x) = |x| \end{align*}[/math] to be continuous at [math]\displaystyle \begin{align*} x = 0 \end{align*}[/math] you need to show [math]\displaystyle \begin{align*} |x - 0| < \delta \implies \left| |x| - 0 \right| < \epsilon \end{align*}[/math]. Working on the second equality we have

[math]\displaystyle \begin{align*} \left| |x| - 0 \right| &< \epsilon \\ \left| | x | \right| &< \epsilon \\ |x| &< \epsilon \\ |x - 0 | &< \epsilon \end{align*}[/math]

So that means we can let [math]\displaystyle \begin{align*} \delta = \epsilon \end{align*}[/math] and reverse the process, this will complete the proof :)

The right approach is the following:

We have shown that:

[math] |a_{n}-2|<\frac{2}{n+1}[/math].

But [math]\frac{2}{n+1}<\frac{2}{n}[/math]

So if we want to choose an N such that for every n>N [math]|a_{n}-2|[/math] is less than epsilon

We simply choose [math] N>\frac{2}{\epsilon}[/math] and

Hence for every n>N WE HAVE:

[math]n>\frac{2}{\epsilon}[/math] and thus:

[math] |a_{n}-2|<\epsilon[/math]

 

[Prove It]

There is nothing wrong with what I posted, if you have the sequence [math]\displaystyle \begin{align*} \frac{2n+3}{n+1} \end{align*}[/math], then it will have the same limit as the function [math]\displaystyle \begin{align*} f(x) = \frac{2x+3}{x + 1} \end{align*}[/math]. So proving the limit for the continuous function will prove the limit for the sequence also.