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I tried solving the question in the title as follows:

$$\lim_{x\to \infty} \frac{4x^2}{x-2} - 4x \to 4x - 4x \to 0$$

However, apparently that first step ($\to 4x - 4x$) was wrong, and I should first have brought the second $4x$ into the numerator.

My question is not how I need to solve the question, as I know that now. My question is why what I did was wrong, as I lack any intuition for it, and it seems a mystery to me.

Erik Gerrits
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  • Why do you think you could replace the first term with $4x$? (At least, this appears to be what you've done, but I'm not sure.) – Theoretical Economist Dec 26 '16 at 08:57
  • I would say $\lim_{x\to \infty} \left(\frac{4x^2}{x-2} - 4x\right)=8$ – georg Dec 26 '16 at 08:59
  • @TheoreticalEconomist, I thought that approaching infinity the $-2$ would be irrelevant, and then I just divided $4x^2$ by $x$. Apparently you cannot do that. But I don't have an intuition for why this is the case. – Erik Gerrits Dec 26 '16 at 09:24
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    If you were to calculate the limit of $\frac{4x^2/(x-2)}{4x}$, your approach, while not technically correct, would've given the right answer $1$ because the $-2$ becomes less and less important. However, in this case (since $\frac{4x^2}{x-2}$ and $4x$ both diverge to infinity), subtraction is much more sensitive to such small differences than division is. To see this more clearly, take a simpler example, like $\lim \frac{x+1}{x}$ vs $\lim[ (x+1)-x]$. – Arthur Dec 26 '16 at 09:26

5 Answers5

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As $x \to \infty$, $\frac{4x^2}{x-2}$ and $4x$ are asymptotically equivalent. However, the notion of asymptotic equivalence is of relative equivalence (in the sense that their ratios tend to $1$). We cannot deduce anything about the differences. The differences may be fixed (e.g. $x^2 \sim x^2 + 1$), tend to $0$ (e.g. $x^2 \sim x^2 + \frac1x$) or tend to infinity (e.g. $x^2 \sim x^2 + x$).

Therefore when working with limits which involve differences, the asymptotic equivalence becomes essentially irrelevant. This is why what you're doing is incorrect.

MathematicsStudent1122
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What yo did was wrong for at least three reasons:

(1) Why would $\;\lim_{x\to\infty}\frac{4x^2}{x-2}=4x\;?$

(2) You can't take the limit when $\;x\to\text{whatever}\;$ and still remain with $\;x\;$ in the final expression

(3) You can't pass to the limit in only part of the expression, which is what ou did here: you let $\;4x\;$ untouched and pretended to calculate the first summand's limit (also this wrongly, as noted in (1)) .

DonAntonio
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Hint

You can use long division first and get $$\frac{4x^2}{x-2}=4 x+8+\frac{16}{x}+\cdots$$

The other way could be $$\frac{4x^2}{x-2}=4\frac{x^2}{x-2}=4\frac{x^2-4x+4+4x-4}{x-2}=4\frac{(x-2)^2+4(x-1)}{x-2}=4\left(x-2+4\frac{x-1}{x-2} \right)=4\left(x-2+4\frac{x-2+1}{x-2} \right)=4\left(x-2+4\left(1+\frac{1}{x-2}\right) \right)=$$ $$4\left(x+2+\frac 4{x-2}\right)=4x+8+\frac {16}{x-2}$$

Claude Leibovici
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In general if you have a set $A \subset \mathbb{R}$ and two functions $f , g : A \to \mathbb{R}$ with $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = \infty$ you can't say that $\lim_{x \to \infty} (f - g)(x) = 0$; for example, take $f(x) = x^2$ and $g(x) = x$.

joseabp91
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You cannot write "$4x^2/(x-2)\to 4x$ as $x\to \infty$ " as it is gibberish: Check the definition of a limit.

In many cases $A(x)/B(x)-C(x)$ may have a limit when neither $A(x)/B(x)$ nor $C(x)$ does.

In your Q, it is worthwhile to put the expression into a form with a common denominator and see what you get. You get $8x/(x-2)$ which can be seen to be equal to $8+16/(x-2) ,$ which converges to $8$ as $x\to \infty.$

DanielWainfleet
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