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Could someone explain where and why this "proof" falls apart?

$\lim_{x\to\infty} x = \infty = \lim_{x\to\infty} x - 1$ implies

\begin{align*} 0 &= \lim_{x\to\infty} 0 \\ &= \lim_{x\to\infty} (x - x) \\ &= \lim_{x\to\infty}x - \lim_{x\to\infty} x \\ &= \lim_{x\to\infty} x - \lim_{x\to\infty} (x-1) \\ &= \lim_{x\to\infty} (x-(x-1)) \\ &= \lim_{x\to\infty} 1 \\ &= 1 \end{align*}

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    $\begingroup$ $ \lim_{x\to\infty}x - \lim_{x\to\infty} x \neq \lim_{x \to \infty} (x-x)$ $\endgroup$ – Sahiba Arora Aug 3 '17 at 21:55
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The identity $$\lim(f(x)+g(x))=\lim f(x)+\lim g(x)$$ which you use from line 2 to line 3, is only valid when all those three limits (or at the very least the two on the right) exist and are finite.

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It fails here: $$\lim_{x\to\infty}(x-x)=\lim_{x\to\infty}x-\lim_{x\to\infty}x.$$ This kind of arithmetic is correct if at least one limit is finite, or minuend and subtrahend tend to the other infinities (differeing in signs). The $$\infty-\infty$$ is undefined form.

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Line 3: $\infty - \infty$ doesn't mean anything.

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