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I have the following definition for a limit:

Definition: Given a function $f : D \rightarrow \mathbb{R}$ and a limit point $x_{0}$ of its domain $D$, for a number $\ell$, we write

$$ \lim_{x\to x_{0}} f(x) = \ell$$

provided that whenever $\{x_{n}\}$ is a sequence in $D \ - \{x_{0}\}$ that converges to $x_{0}$,

$$\lim_{n\to\infty} f(x_{n}) = \ell. $$

Using this definition, I want to show that $\lim_{x\to 0} x^2/x = 0$. Here is my attempt:

Let $\{x_{n}\}$ be a sequence in $\mathbb{R} - \{0\}$ such that $\{x_{n}\}$ converges to $0$. This means for all $\epsilon > 0$, there exists an index $N$ such that

$$|x_{n} - 0| < \epsilon $$

for all $n \geq N$. To prove the original claim, we need to show for all $\epsilon > 0$, there is an index $N'$ such that

$$|\frac{x_{n}^{2}}{x_{n}} - 0| < \epsilon $$

for all $n \geq N'$. But, note that

$$|\frac{x_{n}^{2}}{x_{n}} - 0| = |\frac{x_{n}^{2}}{x_{n}}| = |x_{n}| = |x_{n} - 0|,$$

so setting $N' = N$ suffices. $\blacksquare$

Is my proof correct? Is there anything that can be made better?

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  • $\begingroup$ This proof looks perfect to me. I don't think it needs anything. $\endgroup$
    – Yanko
    Oct 21 '18 at 21:12
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It's correct. You could perhaps be a little more terse, but all and all it's good :)

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It's correct. But if you wanted to be more direct simply noting that for $x_n \ne 0$ then $\frac {x_n^2}{x_n} = x_n$. So $\{x_n\}$ and $\{\frac {x_n^2}{x_n}\}$ are the exact same sequence. And by definition, $x_n \to 0$ so $\frac {x_n^2}{x_n} = x_n \to 0$. By definition.

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That said and done, YOUR answer is better as it is thorough and makes certain that you, the student, are perfectly capable of proving thing specifically by definitions. (Which is essential that all mathematics are capable of.)

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