1. Jan 23, 2016

What is the $\lim_{x \to \infty} x^2$?

What I get is:

$\lim_{x \to \infty} x^2$
$= \lim_{x \to \infty} \frac{\frac{1}{x^2}}{\frac{1}{x^2}} x^2$
$= \lim_{x \to \infty} \frac{\frac{x^2}{x^2}}{\frac{1}{x^2}}$
$= \lim_{x \to \infty} \frac{1}{\frac{1}{x^2}}$
$= \frac{1}{\frac{1}{\infty}}$
$= \frac{1}{0}$

2. Jan 23, 2016

### Staff: Mentor

It looks like you have shown (in a somewhat roundabout way) that the limit is undefined--in other words, that $x^2$ increases without bound as $x$ increases without bound (or, in the more usual sloppy terminology, $x^2 \rightarrow \infty$ as $x \rightarrow \infty$). Does that seem reasonable to you?

3. Jan 25, 2016

### Staff: Mentor

Going from this step to the next, you are saying that $\lim_{x \to \infty} x^2 = \infty$. In other words you have used a number of unnecessary steps to arrive at pretty much the same thing as you started with.

Also, you should never write either $\frac 1 {\infty}$ or $\frac 1 0$. In the first, $\infty$ is not a number that can be used in arithmetic expressions, and in the second, $\frac 1 0$ is undefined.
Much more simply, if x grows large without bound, $x^2$ does the same even more rapidly.

4. Jan 25, 2016

### Staff: Mentor

The common ε-δ-definition of limits turns in the case of unlimited sequences to
$∀ r ∈ ℝ ∃ N ∈ ℕ ∀ n ≥ N : x_n > r$ for $\lim_{n→∞} x_n = ∞$ ($x_n < r$ for $\lim_{n→∞} x_n = -∞$)