# Very simple. What is wrong with this chain of limit equalities?

I know the first line is true and the last line is false. I don't know why the reasoning in between is wrong. $\lim_{x \rightarrow \infty} x^2 = \lim_{x \rightarrow \infty} x$ (True)

$((\lim_{x \rightarrow \infty} x^2)/(\lim_{x \rightarrow \infty} x)) = 1$

$\lim_{x \rightarrow \infty} (x^2/x) = 1$

$\lim_{x \rightarrow \infty} x = 1$ (false)

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Please, state things clearly. Is it $\displaystyle \lim_{x\rightarrow \infty}=c$, where $c$ is a real number, or is it $\displaystyle \lim_{x\rightarrow c}=\infty$?, where $c \in [-\infty, + \infty]$? –  Aloizio Macedo Feb 24 '13 at 22:19

To get to the 2nd line, you "divide both sides by infinity". That's a no-no.

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Yes, thank you. I certainly recognize dividing by infinity is not valid. I had to see that the limit operation is performed (giving infinity) before the division operation is performed. Thanks! –  Just Some Old Man Feb 24 '13 at 22:26

Look at $\lim_{x \rightarrow \infty}(x^2/x)$. Cannot this be reduced? To $\lim_{x \rightarrow \infty}(x/1) = \lim_{x \rightarrow \infty}(x)$. Then its obvious that it does not tend to $1$ for the same reason the last one doesn't.

Oh, sorry, I misunderstood your question. I thought these were four different statements, not four steps to a rearrangement.

As the previous respondent said, you cannot divide infinity by infinity. This is an indeterminate form.

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To operate with limits, is actually to operate with the result of the limit, then you can only manipulate limits as numbers when the limit exists, that is to say, its not $\infty$ or $-\infty$.

So you can't actually get from the first to the second line, because $\lim_{x \rightarrow \infty}x^2=\lim_{x\rightarrow \infty} x = \infty$.

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