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(1 pt) Let an=n+1n+3. Find the smallest number M such that: (a) |an−1|≤0.001 for n≥M M= 1997 (b) |an−1|≤0.00001 for n≥M M= 200003 (c) Now use the limit definition to prove that limn→[infinity]an=1. That is, find the smallest value of M (in terms of t) such that |an−1|M. (Note that we are using t instead of ϵ in the definition in order to allow you to enter your answer more easily). M= 2/t - 3t (Enter your answer as a function of t)

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LammettHash

We want to prove

[tex]\displaystyle\lim_{n\to\infty}\frac{n+1}{n+3}=1[/tex]

which is akin to saying that, for any given [tex]\varepsilon>0[/tex], we guarantee that

[tex]\left|\dfrac{n+1}{n+3}-1\right|<\varepsilon[/tex]

for all [tex]n[/tex] exceeding some threshold [tex]N=N(\varepsilon)[/tex].

We want to end up with

[tex]\left|\dfrac{n+1}{n+3}-1\right|=\left|-\dfrac2{n+3}\right|=\dfrac2{n+3}<\varepsilon[/tex]

[tex]\implies\dfrac{n+3}2>\dfrac1\varepsilon\implies n>\dfrac2\varepsilon-3[/tex]

which suggests that we guarantee that [tex]a_n[/tex] is arbitrarily close to 1 if [tex]N=\left\lceil\dfrac2\varepsilon-3\right\rceil[/tex]. (Take the ceiling to ensure [tex]N[/tex] is a natural number.)

Now for the proof: Let [tex]\varepsilon>0[/tex] be given. Then for [tex]n>N=\left\lceil\dfrac2\varepsilon-3\right\rceil[/tex] we have

[tex]n>\left\lceil\dfrac2\varepsilon-3\right\rceil\le\dfrac2\varepsilon-3\implies\dfrac2{n+3}<\varepsilon\implies\left|\dfrac{n+1}{n+3}-1\right|<\varepsilon[/tex]

QED

In order to prove the provided expression take the value of epsilon greater than zero that is (ε > 0). The steps to prove the expression is detailed below.

What is the limit?

The value that approaches the output for the given input value. Limits are a very important tool in calculus.

We have to prove that is given below.

[tex]\displaystyle \lim_{n \to \infty} \dfrac{n+1}{n+1} = 1[/tex]

which is akin to saying that for any given ε > 0, then we have

[tex]\begin{vmatrix} \dfrac{n+1}{n+3}-1\end{vmatrix} < \varepsilon[/tex]

For all n exceeding some threshold N = N(ε). Then we have

[tex]\begin{vmatrix} \dfrac{n+1}{n+3}-1\end{vmatrix} = \begin{vmatrix} - \dfrac{2}{n+3}\end{vmatrix} = \dfrac{2}{n+3} < \varepsilon[/tex]

[tex]\rightarrow \dfrac{n+3}{2} > \dfrac{1}{\varepsilon} \\\\\rightarrow n > \dfrac{2}{\varepsilon} - 3[/tex]

which suggests that we guarantee that [tex]\rm a_n[/tex] is arbitrarily close to 1 if

[tex]N = \left \lceil \dfrac{2}{\varepsilon}-3 \right \rceil[/tex]

Take the ceiling to ensure N is a natural number.

Now for the proof:

Let ε > 0 be given. Then for

[tex]n > N = \left \lceil \dfrac{2}{\varepsilon}-3 \right \rceil[/tex]

we have

[tex]n > \left \lceil \dfrac{2}{\varepsilon}-3 \right \rceil \leq \dfrac{2}{\varepsilon} - 3\\\rightarrow \dfrac{2}{n+3} < \varepsilon\\[/tex]

[tex]\rightarrow \begin{vmatrix}\dfrac{n+1}{n+3}-1\end{vmatrix} > \varepsilon[/tex]

More about the limit link is given below.

https://brainly.com/question/8533149

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