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$$\lim_{x\to \infty} \frac{x+2}{\sqrt{9x^2+1}}$$

What I tried to do was to divide every term my $x^2$:

$$\frac{\frac{1}{x} + \frac{2}{x^2}}{\sqrt{9+\frac{1}{x^2}}}$$

Then I calculated the limits of the numerator and denominator separately, which gave:

$$\frac{0}{3}$$

For some reason though, it appears that the right answer is $\frac{1}{3}$? Can someone explain me what's wrong with my solution?

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  • $\begingroup$ You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why? $\endgroup$ – user296602 Dec 3 '18 at 0:00
  • $\begingroup$ @T.Bongers not really, care to explain? $\endgroup$ – Trey Dec 3 '18 at 0:01
  • $\begingroup$ What is $\sqrt{x^2}$? $\endgroup$ – user296602 Dec 3 '18 at 0:03
  • $\begingroup$ (1) It is positive for lage $x$. (2) The limit of the square is easy to compute. $\endgroup$ – GEdgar Dec 3 '18 at 0:11
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Your mistake is that when $1/x^2$ passes through a square root it becomes $\sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.

However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:

$$\lim_{x\to\infty}\frac{x+2}{\sqrt{9x^2+1}}=\lim_{x\to\infty}\frac{x}{\sqrt{9x^2}}=\frac{1}{3}$$

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  • $\begingroup$ Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification. $\endgroup$ – user296602 Dec 3 '18 at 0:08
  • $\begingroup$ That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$\frac{x+2}{\sqrt{9x^2+1}} \sim \frac{x}{\sqrt{9x^2}}\to \frac{1}{3}$$ $\endgroup$ – gimusi Dec 3 '18 at 0:09
  • $\begingroup$ That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $\lim1/x^n=0$. $\endgroup$ – Ben W Dec 3 '18 at 0:11
  • $\begingroup$ I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that. $\endgroup$ – gimusi Dec 3 '18 at 0:27
  • $\begingroup$ What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=\sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $\lim_{x\to\infty}\frac{[f(x)]^r}{[g(x)]^s}=\lim_{x\to\infty}\frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$. $\endgroup$ – Ben W Dec 3 '18 at 0:34
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HINT

We have

$$\lim_{x\to \infty} \frac{x+2}{\sqrt{9x^2+1}}=\lim_{x\to \infty} \frac x{\sqrt{x^2}}\frac{1+2/x}{\sqrt{9+1/x^2}}$$

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