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When dealing with limits, we can do $\infty + \infty = \infty $ or $ \infty \cdot \infty = \infty$. But can I something similar for $\infty - \infty$ or $\frac{\infty}{\infty}$?

I'm asking because I can't calculate this $$\lim_{x \to \infty} \sqrt{x^3+3x} \ -\sqrt{x^4-x^2}$$ I have tried to rationalize it, which would make it $$\frac{x^3+3x-x^4+x^2}{\sqrt{x^3+3x} \ +\sqrt{x^4-x^2}}$$ but I would always end up reaching $\infty - \infty$ or $\frac{\infty}{\infty}$. I'm pretty sure we can't do this, so any advice on how to calculate it?

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    Your notation suggests that you are thinking of $\infty$ as a number. It is not a number and needs to be treated differently. – 2017-01-26
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    Certainly not. Because $\infty+\infty=\infty$ and $\infty-\infty=0$ would imply $\infty+\infty-\infty=\infty=0$. – 2017-01-26
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    You cannot perform $\infty-\infty$ as it is undefined. Rewriting might do the trick, for example $\lim_{x \to \infty} {(2x-x)}=\lim_{x \to \infty} {x}=\infty$. – 2017-01-26
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    what you can do in the expression that you got to, is divide the numerator and denominator by $x^2$ then you may be able to see what happens at infinity – 2017-01-26

4 Answers 4

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$$\lim_{x \to \infty}\left(\sqrt{x^3+3x} \ -\sqrt{x^4-x^2}\right)=\lim_{x \to \infty}x^2\left(\sqrt{\frac1x+\frac3{x^3}} \ -\sqrt{1-\frac1{x^2}}\right).$$

The limit of the right factor is $-1$, but that of $x^2$ does not exist (the limit is $-\infty$ if you prefer).

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No, $\infty-\infty=0$ is not allowed and is very very wrong. For example,

$$\lim_{n\to\infty} (n) - (n-x) = x$$ so, depending on the value of $x$, the limit can be literally any real number

Similarly, $\frac\infty\infty=1$ is just as wrong, as $$\lim_{n\to\infty}\frac{xn}{n} = x$$ and the same as above applies.

In your case, you made the correct first step, now I advise you to divide both expressions by $x^2$.

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Limits that approach $\infty-\infty$, $\dfrac{\infty}{\infty}$, $\dfrac{0}{0}$ are undefined. This means we can't defined them as an actual value. In such cases we convert these type of limits to limits have value in $\mathbb{R}$ or $\pm\infty$.

For limits like $$\lim_{x \to \infty} \sqrt{x^3+3x} \ -\sqrt{x^4-x^2}$$ that $x\to\infty$ alwayes the largest power effects on factors. Main power in first radical is $x^3$ and in the last is $x^4$, so we write $$\lim_{x\to\infty}\sqrt{x^3+3x}\ -\sqrt{x^4-x^2}=x^\frac32-x^2$$ again between $x^\frac32$ and $x^2$, $x^2$ is largest thus limit has value $-\infty^2$ that is $-\infty$.

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Rule of thumb for expressions involving powers (and only powers) of $x$ is that the term with the largest power eventually dominates for $x\rightarrow\infty$. Your $\sqrt{x^3+3x}-\sqrt{x^4-x^2}$ will be eventually dominated by $x^4$, which diverges as $x\rightarrow\infty$.