For an exam, I had to evaluate the following limit:


I evaluated the limit to be $3$ and wolfram confirms that but in order to do that I did the following steps:

\begin{align} \require{cancel} &\lim_{x\to\infty}\frac{\sqrt{9x^6-x}}{x^3+1} \\ =&\lim_{x\to\infty}\frac{3x^3-\sqrt{x}}{x^3+1} \quad\spadesuit \\ =&\lim_{x\to\infty}\frac{x^3(3-\frac{\sqrt{x}}{x^3})}{x^3(1+\frac{1}{x^3})} \\ =&\lim_{x\to\infty}\frac{\cancel{x^3}(3-\cancelto{0}{\frac{\sqrt{x}}{x^3}})}{\cancel{x^3}(1+\cancelto{0}{\frac{1}{x^3}})} \\ =& 3 \end{align}

Now, the main assumption I made was in $\spadesuit$ where I assumed that since $x\rightarrow \infty$, the square root must be positive. Is it correct to make that assumption?

Thanks a bunch!

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If $\displaystyle\sqrt{a-b}= \sqrt a-\sqrt b,$

Squaring we get $\displaystyle a-b=a+b-2\sqrt{ab}\iff 2\sqrt{ab}=2b$ which is true iff $\sqrt b=0\iff b=0$ or $\sqrt a=\sqrt b\iff a=b$ assuming $a,b\ge0$

Just divide the numerator & the denominator by $x^3$

or set $\displaystyle\frac1x=h$

Clearly, $\displaystyle x\to+\infty\implies h\to0^+$ and


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  • $\begingroup$ So, I was wrong to make that assumption right? $\endgroup$ – Jeel Shah Jan 6 '14 at 15:36
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    $\begingroup$ No you were wrong when you say that $\sqrt{9x^6-x}=3x^3-\sqrt{x}$. $\endgroup$ – user37238 Jan 6 '14 at 15:37
  • $\begingroup$ @Listing, sorry I implicitly assumed that $a\ne b$ $\endgroup$ – lab bhattacharjee Jan 6 '14 at 17:34
  • $\begingroup$ @gekkostate: I think you missed user37238's point -- you can't distribute a power over each of the terms in a sum. $\endgroup$ – MPW Jan 6 '14 at 18:23
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    $\begingroup$ @Aðøbe, $$\lim_{x\to\infty}\frac{\sqrt{9x^6-x}}{x^3+1}=\lim_{h\to0^+}\frac{\sqrt{\frac9{h^6}-\frac1h}}{\frac1{h^3}+1}$$ $$=\lim_{h\to0^+}\frac{\sqrt{\frac{9-h^5}{h^6}}}{\frac1{h^3}+1}=$$ $\endgroup$ – lab bhattacharjee Jan 6 '14 at 18:56

Just use asymptotic! (which are basically taylor series of the first order)

This way you get $$9x^6 - x \sim 9x^6 \Rightarrow \sqrt{9x^6 - x} \sim \sqrt{9x^6} = 3x^3$$

Plus, $$x^3 + 1 \sim x^3$$, so the limit becomes $$\frac{3x^3}{x^3} = 3$$

(Yes, asymptotics are awesome)

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The numerator for large $x$ "looks like" 3x^3 since the $x^6$ severely dominates the $x$ term in the square root, so the idea is that we want to factor this term out like so:

$$\frac{\sqrt{9x^6-x}}{x^3+1} = \frac{3x^3\sqrt{1-\frac{1}{9x^5}}}{x^3+1} = \frac{3x^3\sqrt{1-\frac{1}{9x^5}}}{x^3\left(1+\frac{1}{x^3}\right)}.$$

Notice that the $x^3$ terms cancel. The remaining limit is straightforward to evaluate.

Your mistake of writing $\sqrt{9x^6-x} = \sqrt{9x^6}-\sqrt{x}$ is a common one. Just like we don't say $(x+y)^2 = x^2+y^2$, we don't say $\sqrt{a+b} = \sqrt{a}+\sqrt{b}$. Powers are not distributive like this.

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