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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 $2.\frac{1}{\infty}=2.0=0$

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    $\begingroup$ As long as you know what you are doing. $\endgroup$ – Megadeth Nov 2 '17 at 23:08
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    $\begingroup$ Only if $r$ is positive. $\endgroup$ – Franklin Pezzuti Dyer Nov 2 '17 at 23:09
  • $\begingroup$ @Nilknarf Yeah thanks, I forgot to mention that. $\endgroup$ – Hai Nov 2 '17 at 23:09
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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 $r>0$.

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

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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 $.

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