# Show the limit of a sequence exists

I'm trying to get some extra practice in before my final exam in two weeks and I ran across this problem. How can we do it?

Part A: Suppose for some $b>0, f: [b, \infty ) \rightarrow R$ is such that $\lim_{x\to \infty} f(x) = L$ for some real number $L$. Let $a_n = f(n)$ for all integers $n>b$. Show that we also have $\lim_{n\to \infty} a_n = L$.

Part B: Give an example to that the converse to the result in part (a) is not true.

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## 3 Answers

Suppose $\lim_{x\to\infty} f(x)=L$. Then for every $\epsilon>0$ there is a $c>b$ such that for all $x>c$ we have $|f(x)-L|<\epsilon$. In particular for all integers $n>c$, we have $|f(n)-L|<\epsilon$.

To show the converse is not true, let $f$ be the function that sends each integer to $1$ and everything else to $0$.

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One alternative definition of the limit is

$$\lim_{x \to a}f(x)=L \Leftrightarrow \forall x_n \to a \text{ we have } f(x_n) \to a$$

This solves part a) because $x_n=n \to \infty$.

One counterexample for the second part can be taken as $f(x)=\sin(2\pi x)$. The function $f$ does not have a limit as $x \to \infty$ because it is oscilatory. Still $f(2\pi n)=0 \to 0$ as $n \to \infty$.

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Part a: $\displaystyle \lim_{x \rightarrow \infty} f(x) = L$ means that $\forall \epsilon > 0$ $\exists X$ s.t. $x \geq X \implies |f(x)-L| < \epsilon$. Hence set $N$ to be the least natural number larger than $X$ and $\forall n \geq N$, $|a_n - L| = |f(n)-L| < \epsilon$

For part B, let $f(x) = \begin{cases} 1 &\mbox{if } x \in \mathbb{N} \\ x &\mbox{otherwise} \end{cases}$ And we're done.

We can even do better if we want $f$ to be continuous and bounded, by saying $f(x) = \sin(\pi x)$

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