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I want to find the $$\lim_{x\rightarrow \infty} (\sqrt x-x)$$

What I was thinking was to multiply by the conjugate since that's normally what you do with square roots, So I would get

$$\lim_{x\rightarrow\infty} (\sqrt{x}-x)\cdot\frac{(\sqrt{x}+x)}{(\sqrt{x}+x)}$$

$=\lim_{x\rightarrow\infty} \frac{x-x^2}{\sqrt{x}+x} = \lim_{x\rightarrow\infty}\frac{x(1-x)}{\sqrt{x} (1+x)} = \lim_{x\rightarrow\infty} \frac{-x}{\sqrt{x}}$

I guess as x goes to infinity, $\sqrt \infty$ is smaller than $-\infty$, so it blows up to $-\infty$ but I'm unsure if this logic is correct.

One solution I found was $\lim_{x\rightarrow\infty} \sqrt x (1-\sqrt x) = (\infty) (1-\infty) = (\infty)(-\infty) = -\infty$

However I'm just uncomfortable actually plugging in $\infty$ for limit values and I also wouldn't think to do this. I would think to multiply by the conjugate

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  • $\begingroup$ $\sqrt{x}-x = x(\frac{1}{\sqrt{x}}-1) \to -\infty$ $\endgroup$ – Jakobian Jun 15 '19 at 23:30
  • $\begingroup$ $\sqrt{x}-x=-\sqrt{x}\left(\sqrt{x}-1\right)\to-\infty$ $\endgroup$ – robjohn Jun 16 '19 at 7:42
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One thing you can also do, which is often looked over, is substituting x.

For example by $f(x)= x^2$. This is allowed, as both $f(x)$ and $x$ tend to $\infty$ for $x\to\infty$.

With this the problem becomes: $$ \lim_{x\rightarrow \infty} (\sqrt x-x) =\lim_{x\rightarrow \infty} (x-x^2) $$

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  • $\begingroup$ thank you! is this kind of like squeeze theorem? $\endgroup$ – user477465 Jun 16 '19 at 0:28
  • $\begingroup$ @user477465 This follows directly from the fact, that both $f(x):=x$ and $f(x):=x^2$ have $\lim_{x\to\infty} f(x)= \infty$. If you take the definition apart, it means basically, that no matter how high an $\epsilon$ you choose, both functions still will eventually reach a point from which they'll grow above it. $\endgroup$ – Sudix Jun 16 '19 at 0:53
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Your intuition is correct even if your manipulations do not literally make sense. Observe that for all $x\geq 2$, we have that $\sqrt{x}\leq x/4$. Thus, $$ \sqrt{x}-x\leq \frac{x}{4}-x=-\frac{3x}{4}, $$ which tends to $-\infty$ as $x\to\infty$. Thus the limit is $-\infty$, by the comparison theorem for limits.

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If $x \geq 1$ then $\frac {x^{2}-x} {\sqrt x +x} \geq \frac {x^{2}-x} {x +x} =\frac 1 2 (x-1) \to \infty$. Now multiply by $-1$.

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