How do you Compute a Squareroot limit?

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Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt\[n\]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$

I am trying to compute the limit here, i am not sure how to do it if i just plug it in or do i do the conjugate and solve? For some reason i got infinity as my answer yet I was told theres an actual solution to it. Perhaps someone can help me out here?

$\lim\limits_{x\rightarrow \infty}\sqrt{16x^2-5x+10{}}-4x$

I think there is an abstract duplicate for this one which I am unable to find out now. If it is not there, I think it would be good to create one. — , Jun 22 '12 at 18:15
@PeterTamaroff Could you then link this to the old one and vote to close it as duplicate? — , Jun 22 '12 at 18:21
I dont think there is, maybe a similar style question but not mine — soniccool, Jun 22 '12 at 18:23

score 2 · Accepted Answer · edited Jun 22 '12 at 18:28

To paraphrase Tom Lehrer: Rationalize, rationalize, rationalize.

We use the fact that $(a-b)(a+b) = a^2-b^2$. So if we multiply $$\sqrt{16x^2-5x+10} - 4x$$ by $$\sqrt{16x^2-5x+10}+4x$$ that will eliminate the square root. Unfortunately, we are not allowed to just multiply by whatever we please (that changes the function), so we must be sure that what we do, in total, does not amount to anything. So in addition to multiplying by $\sqrt{16x^2-5x+10}+4x$, we also divide by it so that it all amounts to multiplying by $1$ (which doesn't do anything). Then we can do the limit the usual way, by noting that the highest degree of $x$ that shows up in both numerator and denominator is $x$. (I know it looks like an $x^2$ in the denominator, but it is inside a square root, so it amounts to just $x$): $$\begin{align*} \lim_{x\to\infty}\left(\sqrt{16x^2-5x+10}-4x\right) &= \lim_{x\to\infty}\frac{(\sqrt{16x^2-5x+10}-4x)(\sqrt{16x^2-5x+10}+4x)}{\sqrt{16x^2-5x+10}+4x}\\ &= \lim_{x\to\infty}\frac{16x^2-5x+10 - (4x)^2}{\sqrt{16x^2-5x+10}+4x}\\ &= \lim_{x\to\infty}\frac{-5x+10}{\sqrt{16x^2-5x+10}+4x}\\ &=\lim_{x\to\infty}\frac{\frac{1}{x}(-5x+10)}{\frac{1}{x}(\sqrt{16x^2-5x+10}+4x)}\\ &= \lim_{x\to\infty}\frac{-5 + \frac{10}{x}}{\frac{1}{x}\sqrt{16x^2-5x+10}+4}\\ &= \lim_{x\to\infty}\frac{-5+\frac{10}{x}}{\sqrt{\frac{16x^2-5x+10}{x^2}}+4}\\ &= \lim_{x\to\infty}\frac{-5+\frac{10}{x}}{\sqrt{16 - \frac{5}{x}+\frac{10}{x^2}} + 4}\\ &= \frac{-5+0}{\sqrt{16-0+0}+4}\\ &= \frac{-5}{4+4}\\ &= -\frac{5}{8}. \end{align*}$$

score 1 · Answer 2 · answered Jun 22 '12 at 18:11

$$\lim_{x\rightarrow \infty}\sqrt{16x^2-5x+10{}}-4x$$

$$\lim_{x\rightarrow \infty}\left(\sqrt{16x^2-5x+10{}}-4x\right) \frac{\sqrt{16x^2-5x+10{}}+4x}{\sqrt{16x^2-5x+10{}}+4x}$$

$$\lim_{x\rightarrow \infty}\left(\sqrt{16x^2-5x+10{}}^2-(4x)^2 \right) \frac{1}{\sqrt{16x^2-5x+10{}}+4x}$$

$$\lim_{x\rightarrow \infty}\left(16x^2-5x+10-16x^2 \right) \frac{1}{\sqrt{16x^2-5x+10{}}+4x}$$

$$\lim_{x\rightarrow \infty}\frac{10-5x}{\sqrt{16x^2-5x+10{}}+4x}$$

Can you move on?

Yeah im trying to rework it i think maybe my algebra was wrong. — soniccool, Jun 22 '12 at 18:13

score 1 · Answer 3 · answered Jun 22 '12 at 18:53

1

put $y=1/x$, then your problem becomes $lim_{y \to 0+}$ $ \frac{\sqrt{10y^2-5y+16}-4}{y}$ which you can solve using L'Hôpital's rule.

answered Jun 22 '12 at 18:53

Aang

14,672

I am not aware of this rule yet right now im trynigto go basic steps – soniccool Jun 22 '12 at 18:59

score 0 · Answer 4 · answered Jun 22 '12 at 18:25

There are alternatives to rationalizing. You can pull a factor of $4x$ out of the square root:

$$\sqrt{16x^2-5x+10}-4x=4x\left(\sqrt{1-\frac5{16x}+\frac5{8x^2}}-1\right)\;.$$

Then you can rewrite the limit as

$$\begin{align*} \lim_{x\to\infty}\Big(\sqrt{16x^2-5x+10{}}-4x\Big)&=4\lim_{x\to\infty}x\left(\sqrt{1-\frac5{16x}+\frac5{8x^2}}-1\right)\\ &=4\lim_{x\to\infty}\frac{\sqrt{1-\frac5{16x}+\frac5{8x^2}}-1}{1/x}\;, \end{align*}$$

at which point it succumbs easily to l’Hospital’s rule.

How do you Compute a Squareroot limit?

4 Answers4