# Manipulating $\lim\limits_{x \to 0}{\frac{\sqrt{x+\sqrt{x}}}{x^n}}$

For $x \to 0$ determine the order of smallness relative to $x$ of the function $\sqrt{x+\sqrt{x}}$.

I need to find $n$ for $\lim\limits_{x \to 0}{\frac{\sqrt{x+\sqrt{x}}}{x^n}} \neq 0$.

Here's what I've got so far:

$\lim\limits_{x \to 0}{\frac{\sqrt{x+\sqrt{x}}}{x^n}}$ $= \lim\limits_{x \to 0}{\sqrt{\frac{x+\sqrt{x}}{x^{2n}}}}$ $\longrightarrow \lim\limits_{x \to 0}{\frac{x+\sqrt{x}}{x^{2n}}}$ $= \lim\limits_{x \to 0}{\frac{x+\sqrt{x}}{x^{2n}}\frac{x-\sqrt{x}}{x-\sqrt{x}}}$ $= \lim\limits_{x \to 0}{\frac{x^2-x}{x^{2n}(x-\sqrt{x})}}$ $= \lim\limits_{x \to 0}{\frac{x-1}{x^{2n-1}(x-\sqrt{x})}}$

At this point, I'm not sure how to manipulate the fraction. Any hints?

Once I finish manipulating the limit to the form $\lim\limits_{x \to 0}{\frac{1}{x^{f(n)}}}$, I can find $n$ by solving $f(n) = 0$.

• I think it makes it easier to manipulate if you put $x=t^4$ for a bit. I'm not sure I'd "rationalise the numerator" as you have done, I'd see whether the binomial expansion of the square root helped. Mar 8, 2017 at 7:57
• @ancientmathematician -- Doing as you recommended: $\lim\limits_{x \to 0}{\frac{\sqrt{x+\sqrt{x}}}{x^n}}$ $= \lim\limits_{t \to 0}{\frac{\sqrt{t^4+\sqrt{t^4}}}{t^{4n}}}$ $= \lim\limits_{t \to 0}{\frac{\sqrt{t^4+t^2}}{t^{4n}}}$ $= \lim\limits_{t \to 0}{\sqrt{\frac{t^4+t^2}{t^{8n}}}}$ $\longrightarrow \lim\limits_{t \to 0}{\frac{t^4+t^2}{t^{8n}}}$ $= \lim\limits_{t \to 0}{\frac{t^2+1}{t^{8n-2}}}$. Once again, I'm lost. Mar 8, 2017 at 8:04
• See @Fimpelliziero's answer as expanded. Mar 8, 2017 at 9:24

If $\lim\limits_{x \to 0}{\frac{\sqrt{x+\sqrt{x}}}{x^n}} = c$ for some $c\neq 0$, then $\lim\limits_{x \to 0}{\frac{x+\sqrt{x}}{x^{2n}}} =c^2$. Now, let $\sqrt{x}=t$. We then wish to find $n$ such that

$$\lim_{t\to0}\frac{t^2+t}{t^{4n}}\neq0$$

It's easy to see that $n=\frac14$ will produce a limit of $1$. Now, if $\alpha=4n<1$, we may multiply the numerator and denominator by $t^{-\alpha}$ to obtain that the limit equals

$$\lim_{t\to0}\,\,t^{2-a}+t^{1-\alpha}$$

and since each exponent above is positive, the limit is $0$. It follows that $n=\frac14$ is the smallest $n$ for which the limit is nonzero.

EDIT: Rewriting this answer in terms of ancient mathematician's comment is perhaps simpler. Letting $x=t^4$ in the initial limit, we may rewrite it as

$$L=\lim_{t\to0}\frac{\sqrt{t^4+t^2}}{t^{4n}}=\lim_{t\to0}\,\,t^{1-4n}\cdot\sqrt{t^2+1},$$

and from the first term it's easy to see that $n<\frac14$ implies $L=0$ while $n=\frac14$ implies $L=1$.

• Is verifying that $n$ is the minimum (as done in the main portion) also necessary for the "EDIT" portion? Mar 8, 2017 at 8:13
• Well, it depends on how obvious it seems to you, in both cases really. Mar 8, 2017 at 8:55