26
$\begingroup$

I got a limit:

$$\lim_{x\to\infty}\frac {(2x+3)^3(3x-2)^2} {(x^5 + 5)}$$

As far as $x$ approaches infinity, can I just forget about 'small' numbers (like $3$, $-2$ and $5$ in this example)? I mean is it legal to make a transition to:

$$\lim_{x\to\infty}\frac {(2x)^3(3x)^2} {x^5}$$

Or if it is not always okay — in what cases such transitions are okay?

$\endgroup$
  • 1
    $\begingroup$ Yes it's ok see little oh notation for more information. $\endgroup$ – As soon as possible Nov 4 '17 at 11:11
  • 6
    $\begingroup$ Not only numbers but also smaller powers $$\lim_{x\to\infty}\frac{2x^4+x^3-x^2+4x}{x^4-x^2+1}=\lim_{x\to\infty}\frac{2x^4}{x^4}=2$$ $\endgroup$ – Raffaele Nov 4 '17 at 12:00
  • 25
    $\begingroup$ @ParamanandSingh: I'd refrain from calling it "correct" as opposed to "possible to make rigorous in this particular case". =) $\endgroup$ – user21820 Nov 4 '17 at 12:53
  • 3
    $\begingroup$ @user21820: I meant the same, but your choice of words is better. +1 for your comment. $\endgroup$ – Paramanand Singh Nov 4 '17 at 14:08
  • 2
    $\begingroup$ Yes and no. For example, you can’t just ignore the $2$ in $2x+3$. So you should be more systematic, or rigorous, in your statement of the rule. $\endgroup$ – MPW Nov 4 '17 at 23:30
31
$\begingroup$

As mentioned in the comments, the correct way to make such intuitive arguments rigorous is via asymptotic analysis using Landau notation such as done here and here: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\wi{\subseteq} $

As $x \to \infty$:

  $\lfrac{(2x+3)^3·(3x-2)^2}{x^5+5} = \lfrac{(2+\lfrac3x)^3·(3-\lfrac2x)^2}{1+\lfrac5{x^5}} \in \lfrac{(2+o(1))^3·(3+o(1))^2}{1+o(1)}$

  $\ \wi (2+o(1))^3·(3+o(1))^2·(1+o(1)) \wi (2^3+o(1))·(3^2+o(1))·(1+o(1))$

  $\ = 2^3·3^2·1+o(1) \to 72$.


Note that it is absolutely incorrect to always eliminate small terms in each expression. So it is excellent that you ask your question about when it is valid. Consider the question of finding $\lim_{x \to 0} \lfrac{\exp(x)-1-\sin(x)}{x^2}$ if it exists. If you simply 'eliminate' small terms, then you would get $\lfrac{\exp(0)-1-\sin(0)}{x^2} = 0$, which is not the desired limit. Notice how the proper asymptotic analysis will never fail:

As $x \to 0$:

  $\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{\exp(o(1))-1-\sin(o(1))}{x^2} \wi \lfrac{(1+o(1))-1-o(1)}{x^2} \wi \lfrac{o(1)}{x^2}$.

  [Note that in the last step above you cannot cancel the "$o(1)$" because it is a class of values.]

  [So you are stuck because the final "$\lfrac{o(1)}{x^2}$" is too loose a bound even though it is not wrong.]

  [This tells us that we need more precision in the asymptotic expansion, so we try again.]

  $\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{(1+x+o(x))-1-(x+o(x))}{x^2} \wi \lfrac{o(x)}{x^2}$.

  [Again we get stuck, even though "$\lfrac{o(x)}{x^2}$" is now a tighter bound. So refine more!]

  $\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{(1+x+\lfrac12x^2+o(x^2))-1-(x+o(x^2))}{x^2} \wi \lfrac{\lfrac12x^2+o(x^2)}{x^2} = \lfrac12+o(1) \to \lfrac12$.

  [There we go; we have found the limit, but we can refine further to get even more information!]

  $\lfrac{\exp(x)-1-\sin(x)}{x^2} \in \lfrac{(1+x+\lfrac12x^2+\lfrac16x^3+O(x^4))-1-(x-\lfrac16x^3+O(x^5))}{x^2} \wi \lfrac{\lfrac12x^2+\lfrac13x^3+O(x^4)}{x^2}$

  $\ = \lfrac12+\lfrac13x+O(x^2)$.

$\endgroup$
  • 5
    $\begingroup$ +1. To me this answer demonstrates two things: (1) the power of asymptotic analysis and O notation, and (2) the awkwardness of using set relations for pedantic reasons instead of simply using equality signs. :-) $\endgroup$ – ShreevatsaR Nov 4 '17 at 16:25
  • 3
    $\begingroup$ @ShreevatsaR: Haha yea I'm pedantic. =) I agree it is slightly awkward, but until I find an alternative that is as precise, I think I'll stick to it. Note that I'm already using "$\to$" in a generalized manner so that I can say that entire asymptotic classes of reals tend to a single real. But at least I don't use "$=$" for an asymmetric relation. =) $\endgroup$ – user21820 Nov 4 '17 at 16:38
  • 4
    $\begingroup$ “Aristotle is a man, but a man isn't necessarily Aristotle.” It's interesting how people are willing to accept changes/extensions to the meaning of other notational symbols (for example “$+$” meaning Minkowski sum, as in “$A + B$” or “$x + A$”), but less able to accept an asymmetric “$=$” symbol. (This may have something to do with how primal this verb is… I've read somewhere how some students enter college with $=$ as the only verb in their vocabulary, and given exercises in differentiation, write things like $\sin x = \cos x$.) There may be something interesting here for a psychologist. $\endgroup$ – ShreevatsaR Nov 4 '17 at 18:45
  • 1
    $\begingroup$ This may be correct but looks like overkill, given the tone of the question. $\endgroup$ – Rolazaro Azeveires Nov 4 '17 at 21:08
  • 3
    $\begingroup$ The discussion in comments is as interesting as the answer. $\endgroup$ – Paramanand Singh Nov 5 '17 at 4:15
33
$\begingroup$

Since the question is very broad (it doesn't even mention if the cases you want to consider are always fractions, or if the "small numbers" are constants etc), it might be useful to give a word of warning: try always to do what, for example, @Ennar or @user236182 did in their answer. The "small compared to" logic can fail.

For instance, one might argue that as $x$ goes to $+\infty$, $\sqrt{x^2+x}-x \sim \sqrt{x^2} -x=x-x \to 0$, since $x^2+x \sim x^2$, due to the fact that $x^2$ is the leading term. However, the limit $$\lim_{x \to \infty} \sqrt{x^2+x}-x$$ is not $0$, and may be a good exercise to figure out what it is.

$\endgroup$
  • 2
    $\begingroup$ Excellent example! I've given another example in my answer as well as how to rigorously determine such limits without even knowing the answer. $\endgroup$ – user21820 Nov 4 '17 at 13:25
  • $\begingroup$ Is it (1/2)? I tried solving it. $\endgroup$ – Harry Weasley Nov 4 '17 at 17:17
  • 2
    $\begingroup$ @HarryWeasley Let $x=\frac{1}{t}$. Then $$\lim_{x\to +\infty}\sqrt{x^2+x}-x=\lim_{t\to 0^+}\frac{\sqrt{1+t}-1}{t}=$$ $$=(\sqrt{1+t})'\bigg|_{t=0^+}=\frac{1}{2\sqrt{1+t}}\bigg|_{t=0^+}=\frac{1}{2}$$ Another solution is you can multiply $\sqrt{x^2+x}-x$ by $\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}$ and use the formula $a^2-b^2=(a+b)(a-b)$. In general there's the formula/identity $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1})$$ $k\ge 2$, $k\in\mathbb Z$, $a,b\in\mathbb R$. $\endgroup$ – user236182 Nov 4 '17 at 17:30
  • 4
    $\begingroup$ @HarryWeasley I just factor out the $x^2$ from the square root: $\sqrt{x^2+x}=x\sqrt{1+1/x}$, which, in the limit $x\to+\infty$, can be written as $x(1+1/2x)$ using the standard approximation $\sqrt{1+\epsilon}=1+\tfrac{1}{2}\epsilon$. $\endgroup$ – Viktor Toth Nov 4 '17 at 19:54
  • 2
    $\begingroup$ @ViktorToth: See my above comment! Although I presume you know how to truncate approximations for your desired purposes, I think it's best to make it clear to be precise when explaining to others, and not drop the error terms. =) $\endgroup$ – user21820 Nov 5 '17 at 3:52
4
$\begingroup$

It is always good to specify the steps which allow you to "forget" those numbers, at least roughly: $$\lim_{x \rightarrow \infty} \frac{(2x+3)^3(3x-2)^2}{x^5+5} = \lim_{x \rightarrow \infty}\frac{72x^5+(\text{terms of degree} < 5)}{x^5\Big(1+\frac{5}{x^5} \Big)} = \lim_{x \rightarrow \infty} \frac{72+\frac{(\text{terms of degree }<5)}{x^5}}{1+\frac{5}{x^5}}$$ and now $\frac{\text{terms of degree } < 5}{x^5} \rightarrow 0$ as $x \rightarrow \infty$, as well as $\frac{5}{x^5}$, so you get $72$.

$\endgroup$
3
$\begingroup$

When taking the ratio of polynomials, only the leading terms matter, as they are dominant. So indeed, in the expansion of the factored expression, you can ignore the lower order terms.

The difference of the degree of the numerator and denominator tells you about the limit:

  • $n<d$: $\to 0,$
  • $n=d$: $\to \dfrac{a_n}{b_d}$, where $a_n,b_d$ are the coefficients of the leading terms,
  • $n>d$: $\to \pm\infty$, depending on the sign of $\dfrac{a_n}{b_d}$.
$\endgroup$
1
$\begingroup$

We have $$\lim_{x\to\infty}\frac{2x+3}{2x} = 1,\ \lim_{x\to\infty}\frac{3x-2}{3x} = 1,\ \lim_{x\to\infty}\frac{x^5+5}{x^5} = 1$$

and so $$\lim_{x\to\infty}\frac{(2x+3)^3(3x-2)^2}{x^5+5}= \lim_{x\to\infty}\left(\frac{\frac{(2x+3)^3}{(2x)^3}\cdot\frac{(3x-2)^2}{(3x)^2}}{\frac{x^5+5}{x^5}}\cdot\frac{(2x)^3(3x)^2}{x^5}\right) = \lim_{x\to\infty}\frac{(2x)^3(3x)^2}{x^5}.$$

$\endgroup$
  • 4
    $\begingroup$ This does not answer the question. The question is not about this specific limit, the question is about what transformations are valid when one "simplifies" limit expressions. You don't answer the question in the title -- You don't even acknowledge the question in the title. $\endgroup$ – R.M. Nov 4 '17 at 13:24
  • $\begingroup$ @R.M. I do not agree with you, this clearly demonstrates why ignoring small numbers is valid in cases like these and is a general technique for evaluating lots of limits. To rephrase, this actually gives an elementary approach to asymptotics, which I found more appropriate than discussing asymptotic analysis per se. $\endgroup$ – Ennar Nov 4 '17 at 13:33
  • $\begingroup$ @R.M.: You can see my answer for explicit details of how to properly do such 'transformations'. However, I do agree with Ennar that elementary methods can be used instead of asymptotic analysis, though I still think that asymptotic analysis is the way to go in general. $\endgroup$ – user21820 Nov 4 '17 at 13:46
1
$\begingroup$

$$\frac {(2x+3)^3(3x-2)^2} {x^5 + 5}=$$

$$=\frac{\left(2+\frac{3}{x}\right)^3\left(3-\frac{2}{x}\right)^2}{1+\frac{5}{x^5}}$$

So the limit as $x\to\infty$ is $\frac{(2+0)^3(3-0)^2}{1+0}$.

Edit: one generalization is obvious: to find the limit as $x\to +\infty$ or $-\infty$ of a ratio of two polynomials, divide each term of each polynomial by $x^t$, where $t$ is the degree of the polynomial with the highest degree. Then you can ignore certain terms with smaller degrees:

$$\lim_{x\to +\infty\text{ or }-\infty}\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0}=$$

$$=\lim_{x\to +\infty\text{ or }-\infty}\frac{a_mx^m}{b_nx^n}=\begin{cases}\frac{a_m}{b_n},\ \ \text{ if }m=n\\0,\ \ \text{ if }m<n\\ +\infty\text{ or }-\infty,\ \ \text{ if }m>n\end{cases}$$

I ignored certain terms with smaller degrees.

$\endgroup$
  • 6
    $\begingroup$ This does not answer the question. The question is not about this specific limit, the question is about what transformations are valid when one "simplifies" limit expressions. You don't answer the question in the title -- You don't even acknowledge the question in the title. $\endgroup$ – R.M. Nov 4 '17 at 13:23
  • $\begingroup$ @R.M. I prove that it's ok to ignore "small" numbers in this example and it's obvious how to generalize this. $\endgroup$ – user236182 Nov 4 '17 at 13:31
  • 2
    $\begingroup$ @user236182 It is not obvious to me how to generalize this. Could you add it to your answer? (I'm reminded of the starred exercise in the classic Mathematics Made Difficult which goes “Show that $17 \times 17 = 289$. Generalise this result.”) $\endgroup$ – ShreevatsaR Nov 4 '17 at 16:23
  • 1
    $\begingroup$ @ShreevatsaR One generalization is obvious: to find the limit as $x\to \infty$ of a ratio of two polynomials, divide each term of each polynomial by $x^t$, where $t$ is the degree of the polynomial with the highest degree. $\endgroup$ – user236182 Nov 4 '17 at 16:49
  • $\begingroup$ @user236182 When you say "one generalization", you suggest that there are other generalizations possible, and that there isn't one unique obvious generalization (which is what I said). Anyway, if the generalization you were thinking of is the one you wrote that only applies to limits that are ratios of polynomials, then it is considerably narrower than what the question asks for, so it's worth mentioning in the answer. $\endgroup$ – ShreevatsaR Nov 4 '17 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.