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I have this limit: $$ \lim_{x\to \infty}x^2-x^2\cdot \cos\left(\frac{1}{x}\right) $$

For me my initial answer would be zero as: $$ \lim_{x\to \infty}x^2-x^2\cdot \cos\left(\frac{1}{x}\right)=\lim_{x\to \infty}x^2-\lim_{x\to \infty}x^2\cdot\lim_{x\to \infty}\cos\left(\frac{1}{x}\right) $$ Which is: $$ \infty-\infty\cdot1=0 $$ But after looking at wolfram alpha and doing a series expansion of $\cos(x)$ i see that the answer is in fact $1/2.$

Why is my original thinking incorrect?

MachineLearner
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Alexander Skaane
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    Your calculation is also not valid. You can not calculate with the symbol $\infty$ like that. $\infty-\infty$ is not defined. – Cornman Sep 30 '21 at 19:16
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    Interpreted in terms of limits, it's not always true that $\infty-\infty=0$. Note that $\lim_{x\to\infty}(x+1)=\infty$ and $\lim_{x\to\infty}x=\infty$, but $\lim_{x\to\infty}\left[(x+1)-x\right]=1$. – Alann Rosas Sep 30 '21 at 19:17
  • ok, so am I right in saying that the limit calculation rules do not apply for when x tends to infinity? – Alexander Skaane Sep 30 '21 at 19:20
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    Limit of sum is sum of limits when the limits are defined – J. W. Tanner Sep 30 '21 at 19:20
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    @Cornman $cos(1/x)$ alternates between +1 and -1 as $x$ approaches 0. As $x$ approaches infinity, $1/x$ approaches 0, so don't we just have the limit of $cos(x)$ as $x$ approaches 0? Seems like the cosine limit is just 1. – Nuclear Hoagie Sep 30 '21 at 19:25
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    @Cornman: your last comment disagrees with your first. $\cos(1/x) \to \cos(0) = 1$ as $x \to \infty$. – Rob Arthan Sep 30 '21 at 19:43
  • Note that another way to see that things can be 'funny' here is to factor out the $x^2$ to get $x^2\cdot(1-\cos(1/x))\to\infty\cdot0$ which is one of the canonical 'indeterminate limit' forms. – Steven Stadnicki Sep 30 '21 at 19:44
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    @Cornman I'm not following you - nowhere do we need to take the limit of $cos(1/x)$ as $x$ goes to 0. We need the limit of $cos(1/x)$ as $x$ goes to infinity. Your last statement is true for arbitrary, non-integer $n$, which again shows that the limit of $cos(1/x)$ as $x$ approaches 0 indeed does not exist. But I fail to see why that tells us anything about the limit of $cos(1/x)$ as $x$ approaches infinity. – Nuclear Hoagie Sep 30 '21 at 19:47
  • The real flaw is this: You say that $\infty - \infty = 0$. This is not how limits work at all. If two limits are both infinity, the limit of the difference can be anything at all. – Mark Saving Sep 30 '21 at 20:32

3 Answers3

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Hint: $\cos(x) = 1-\dfrac{x^2}{2}+O(x^4)$ or $\cos(1/x) = 1 - \dfrac{1}{2x^2}+O\left(\dfrac{1}{x^4}\right)$

MachineLearner
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$$\lim_{x\to\infty} x^2(1-\cos(1/x)) = \lim_{x\to\infty} \dfrac{1-\cos(1/x)}{1/x^2} \to {0\over 0}$$

Now you can apply L'Hopital's rule

$$\lim_{x\to\infty} \dfrac{1-\cos(1/x)}{1/x^2} =\lim_{x\to\infty} = \dfrac{-\sin(1/x)/x^2}{-2/x^3} = \dfrac{1}{2} \lim_{x\to\infty} \cdot \dfrac{\sin(1/x)}{1/x}$$

One more application of L'H:

$$\frac{1}{2}\lim_{x\to\infty} \dfrac{\sin(1/x)}{1/x} = \frac{1}{2}\lim_{x\to\infty} \dfrac{\cos(1/x)/x^2}{1/x^2} =\frac{1}{2}\lim_{x\to\infty} \cos(1/x) =\frac{1}{2}\cdot1=\frac{1}{2}$$

David P
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    Suggestion on problems like this: Before applying L'hopitals, do a variable substitution, $y=1/x$, changes the limit to $y\to 0$ and makes the derivatives much nicer – Alan Sep 30 '21 at 19:49
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In general, if you are to calculate $$ \lim_{x \to a} [f(x) + g(x)] $$ where $a$ can be a real number or $\pm \infty$, you cannot blindly move the limit inside the parantheses like this: $$ \Big[\lim_{x\to a} f(x) + \lim_{x \to a} g(x) \Big] $$ The reason you can not do that is because when proving this limit rule it is assumed that both $$\lim_{x\to a} f(x)$$ and $$ \lim_{x \to a} g(x) $$

are defined. To be defined in this context means that they actually are equal to a real number (they converge). If any one of them, say $\lim_{x \to \infty} f(x)$, does not converge to any real number, it could be because for example:

  1. $\lim_{x \to \infty} f(x)$ oscillates between several different values as $x$ goes to $\infty$, and thus does not settle for a limit value. An example of this is the function $\cos(x)$.

  2. $\lim_{x \to \infty} f(x) $ does not have an upper bound. That is, no matter what number $M$ you come up with, there is a point $x_0$ on the real line such that $f(x) > M$ whenever $x > x_0$.

In this latter case we might informally write this as $\lim_{x \to \infty} f(x) = \infty$, but note that this is just notation. $\infty$ is not a number that one can do arithmetic with as with real numbers. What the notation really means is once again that $f(x)$ grows without limit if we only increase $x$ enough. Therefore, an expression such as $\infty - \infty$ does not have any mathematical meaning, because $\infty$ is just a symbol and not a number.

The same caution applies to many other limit rules such as $\lim_{x \to a}(f(x)g(x)) = \lim_{x \to a}f(x) \lim_{x \to a}g(x)$, because you encounter the same problems of doing arithmetic with the symbol $\infty$ otherwise.

One can sometimes convert a limit where the variable goes to $\infty$ to a limit where the variable goes to $0$ by doing a change of variables. In this case, we can make the change of variables $u = \dfrac{1}{2x}$. It is then true that $u \to 0$ when $x \to \infty$. The limit calculation becomes
\begin{align} &\lim_{x\to \infty} \Big[ x^2 - x^2 \cos(\frac{1}{x}) \Big] =\\ &\lim_{u \to 0} \Big[\dfrac{1}{4u^2} - \dfrac{1}{4u^2}\cos(u + u) \Big] = \\ &\lim_{u \to 0} \Big[\dfrac{1}{4u^2} (1 - \cos(u)^2 + \sin(u)^2) \Big] = \\ &\lim_{u \to 0} \dfrac{2 \sin(u)^2}{4u^2} = \\ &\lim_{u \to 0}\dfrac{1}{2} \lim_{u \to 0}\dfrac{\sin(u)}{u} \lim_{u \to 0}\dfrac{\sin(u)}{u} = \\ &\dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{1}{2} \\ \end{align}

Where we have used the facts that

  • $\lim_{u \to 0}\dfrac{1}{2}$ and the standard limit $\lim_{u \to 0}\dfrac{\sin(u)}{u}$ are real numbers.
  • $\cos(u + u) = \cos(u)^2 - \sin(u)^2$
  • $1 = \cos(u)^2 + \sin(u)^2$
Paradox
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