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lim(d^2 --> infty) d - 1 = infty Is this valid notation, or must it be written using lim(d --> infty) instead? I would like to express that as d^2 tends to infinity, d - 1 tends to infinity as well. Is there a better way to write this?

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It sounds to me then, like you are saying $\lim_{d \rightarrow \infty} (\sqrt{d} - 1) = \infty$. –  Jared Apr 13 at 23:05

2 Answers 2

Consider that $d-1 $ is a function of $d^2$. Why sould it be $\sqrt{d^2}-1$ and not $-\sqrt{d^2}-1$?

For the first one, you get the limit $\infty$ and for the second one, $-\infty$.

You should probably change your notation.

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You can write $$\lim_{d\to\pm\infty} d-1=\infty$$ or more accurate $$\lim_{d\to\pm\infty} d-1=\pm\infty$$ or if you want $$\lim_{d^2\to\infty} d-1=\infty$$

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