# Finding the limit with indeterminate form

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

• $\sqrt{x}-x = x(\frac{1}{\sqrt{x}}-1) \to -\infty$ – Jakobian Jun 15 '19 at 23:30
• $\sqrt{x}-x=-\sqrt{x}\left(\sqrt{x}-1\right)\to-\infty$ – robjohn Jun 16 '19 at 7:42

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: $$\lim_{x\rightarrow \infty} (\sqrt x-x) =\lim_{x\rightarrow \infty} (x-x^2)$$
• @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. – Sudix Jun 16 '19 at 0:53
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.
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 $$-1$$.