# Using the sequential definition of a limit to show $\lim_{x\to 0} \frac{x^2}{x} = 0.$

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?

• This proof looks perfect to me. I don't think it needs anything. Oct 21 '18 at 21:12

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.