# Is it valid to write $\lim_{x \rightarrow \infty}\frac{2}{x^r}=2.\frac{1}{\infty}=0$ in limits?

I'm wondering if it's valid to write the follinwg: $$\lim_{x \rightarrow \infty}\frac{2}{x^r}=2\lim_{x \rightarrow \infty}\frac{1}{x^r}=2.\frac{1}{\infty}=2.0=0$$

I know it's valid to say that $\frac{1}{\infty}=0$ in limits but I'm not suring if it would be valid to say $2.\frac{1}{\infty}=2.0=0$

• As long as you know what you are doing. – Megadeth Nov 2 '17 at 23:08
• Only if $r$ is positive. – Franklin Pezzuti Dyer Nov 2 '17 at 23:09
• @Nilknarf Yeah thanks, I forgot to mention that. – Hai Nov 2 '17 at 23:09

Since the limit of a product is the product of the limits: $$\lim_{x\to \infty} \frac{2}{x^r}= 2\lim_{x\to\infty}\frac{1}{x^r}= 2\times 0= 0\, ,\qquad (r>0)\,$$ since $\lim_{x\to \infty}1/x^r=0$ for $r>0$.

You should avoid manipulating $\infty$ like numbers. your result is right just skip the step where you wrote $\frac 1 {\infty}$

There are some operations with infinite limits that are valid. One of them is as follows:

Let $(x_n)_{n \in \mathbb N}$ and $(y_n)_{n \in \mathbb N}$ sequences of positive real numbres such that:

$(x_n)_{n \in \mathbb N}$ is bounded and $\lim_{n \to \infty} y_n = +\infty$

Then

$\lim_{n \to \infty} {x_n}/{y_n} = 0$.

This property remains valid if we consider functions rather than sequences. In this case, the constant function equal to 2 is bounded and the function $x^r$ is such that it tends to infinity where $x$ tends to infinity. Where $r > 0$.