Proof of sequence convergence via the "ε-N" definition

[Euler2718]

Homework Statement

Prove that \lim \frac{n+100}{n^{2}+1} = 0

Homework Equations

(x_{n}) converges to L if
\forall \hspace{0.2cm} \epsilon > 0 \hspace{0.2cm} \exists \hspace{0.2cm} N\in \mathbb{N} \hspace{0.2cm} \text{such that} \hspace{0.2cm} \forall n\geq N \hspace{0.2cm} , |x_{n}-L|< \epsilon

The Attempt at a Solution

I understand most of the workings behind the solution, however there's one step that I can't seem to make a connection with. The solution I am studying goes:

For all \epsilon >0, consider the following inequality:
\left| \frac{n+100}{n^{2}+1}-0\right| = \left| \frac{n+100}{n^{2}+1}\right| < \epsilon
In fact, one has
\left| \frac{n+100}{n^{2}+1} \right| \leq \left| \frac{2}{n}\right|, \hspace{0.2cm} \forall \hspace{0.2cm}n \geq 100

This \frac{2}{n} is causing the problem. I understand how it is used thereafter in the solution, but how they determine this is my question. I think that, for large n, \frac{n+100}{n^{2}+1} is kind of like \frac{n}{n^{2}} = \frac{1}{n}, which is almost correct, but not sufficient enough. I guess the question is, how can I be sure that quantity I deduce - in this case \frac{1}{n}, and in their case \frac{2}{n} - will satisfy (or not satisfy) the inequality at hand. Sometimes it is easy, but I find with more complicated expressions it requires some sort of massaging. Note, I won't have access to a calculator/computer to check in the future.

 

[fresh_42]

There is no need to take care of how big $N=N(\varepsilon)$ is or how close the inequalities are. All which matters is, that you have
$$
\left| \frac{n+100}{n^2+1} \right| < \ldots < \varepsilon
$$
So how to find $N(\varepsilon)$ is not the question. Be generous: $\frac{n+100}{n^2+1} < \frac{2n}{n^2+1} < \frac{2n}{n^2} = \frac{2}{n}$ for $n > 100$. So the $2$ comes in naturally as one substitutes the disturbing $100$ by the larger $n$. So it's not especially tricky but rather especially sloppy.

 

[Mark44]

Homework Statement

Prove that \lim \frac{n+100}{n^{2}+1} = 0

Homework Equations

(x_{n}) converges to L if
\forall \hspace{0.2cm} \epsilon &gt; 0 \hspace{0.2cm} \exists \hspace{0.2cm} N\in \mathbb{N} \hspace{0.2cm} \text{such that} \hspace{0.2cm} \forall n\geq N \hspace{0.2cm} , |x_{n}-L|&lt; \epsilon

The Attempt at a Solution

I understand most of the workings behind the solution, however there's one step that I can't seem to make a connection with. The solution I am studying goes:

For all \epsilon &gt;0, consider the following inequality:
\left| \frac{n+100}{n^{2}+1}-0\right| = \left| \frac{n+100}{n^{2}+1}\right| &lt; \epsilon
In fact, one has
\left| \frac{n+100}{n^{2}+1} \right| \leq \left| \frac{2}{n}\right|, \hspace{0.2cm} \forall \hspace{0.2cm}n \geq 100

This \frac{2}{n} is causing the problem. I understand how it is used thereafter in the solution, but how they determine this is my question.

It's a combination of trial and error along with experience. As you note, $\frac{n + 100}{n^2 + 1}$ is similar to $\frac n {n^2}$ in its long-term behavior, but that's not good enough. Someone made the realization that the first fraction could be made smaller than (or equal to) $\frac 2 n$, and figured out what values of n would make this true.

I think that, for large n, \frac{n+100}{n^{2}+1} is kind of like \frac{n}{n^{2}} = \frac{1}{n}, which is almost correct, but not sufficient enough. I guess the question is, how can I be sure that quantity I deduce - in this case \frac{1}{n}, and in their case \frac{2}{n} - will satisfy (or not satisfy) the inequality at hand. Sometimes it is easy, but I find with more complicated expressions it requires some sort of massaging. Note, I won't have access to a calculator/computer to check in the future.

 

[Euler2718]

There is no need to take care of how big $N=N(\varepsilon)$ is or how close the inequalities are. All which matters is, that you have
$$
\left| \frac{n+100}{n^2+1} \right| < \ldots < \varepsilon
$$
So how to find $N(\varepsilon)$ is not the question. Be generous: $\frac{n+100}{n^2+1} < \frac{2n}{n^2+1} < \frac{2n}{n^2} = \frac{2}{n}$ for $n > 100$. So the $2$ comes in naturally as one substitutes the disturbing $100$ by the larger $n$. So it's not especially tricky but rather especially sloppy.

It's a combination of trial and error along with experience. As you note, $\frac{n + 100}{n^2 + 1}$ is similar to $\frac n {n^2}$ in its long-term behavior, but that's not good enough. Someone made the realization that the first fraction could be made smaller than (or equal to) $\frac 2 n$, and figured out what values of n would make this true.

Ah, I understand. Thank you for this insight.