[Math] Limits and infinity minus infinity

calculuslimits-without-lhopitalradicals

I understand that we need to rationalize when we have infinity minus infinity like here

$\lim_{x\to \infty}\left(\sqrt{x^2 + 1} – \sqrt{x^2 + 2}\right)$

My question is why can I not just split the limits like this

$\lim_{x\to \infty}\left(\sqrt{x^2 + 1}\right) – \lim_{x\to \infty}\left(\sqrt{x^2 + 2}\right)$

and then

$\lim_{x\to \infty}\sqrt{x^2} * \lim_{x\to \infty}\sqrt{1 + \frac 1x} – \lim_{x\to \infty}\sqrt{x^2} * \lim_{x\to \infty}\sqrt{1 + \frac 2x}$

which gives

$\lim_{x\to \infty}\sqrt{x^2} – \lim_{x\to \infty}\sqrt{x^2} = 0$

because $\frac 1x$ and $\frac 2x$ tend to $0$

Where am i wrong ?

Best Answer

You can not subtract $$ \lim_{x\to\infty}(\sqrt{x^2+1}-\sqrt{x^2+2})=\lim_{x\to\infty}\sqrt{x^2+1}-\lim_{x\to\infty}\sqrt{x^2+2}, $$ because the left hand side is $\infty-\infty$ which is not definable.

Instead, you can obtain that $$ \sqrt{x^2+1}-\sqrt{x^2+2}=\frac{(\sqrt{x^2+1}-\sqrt{x^2+2})(\sqrt{x^2+1}+\sqrt{x^2+2})}{\sqrt{x^2+1}+\sqrt{x^2+2}}\\ =\frac{(x^2+1)-(x^2+2)}{\sqrt{x^2+1}+\sqrt{x^2+2}}=-\frac{1}{\sqrt{x^2+1}+\sqrt{x^2+2}}\to 0, $$ where the properties of limits apply.

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