Is it okay to "ignore" small numbers in limits where $x$ approaches infinity?

Question

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?

Answer 1

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.

Answer 2

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!]

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

  $\ = \lfrac12+\lfrac13x+O(x^2)$.

Answer 3

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 $72$.

Answer 4

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 $\dfrac{a_n}{b_d}$.

Answer 5

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

Answer 6

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

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

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.