The limit does exist and it is equal to $\frac{3}{4}$. We get it using L'Hospital. On the other hand I could write the above limit as: $$\lim_{x \to 0} \frac{1-\cos^3 x}{x\sin 2x} = \lim_{x \to 0} \frac{1}{x\sin 2x} - \lim_{x \to 0} \frac{\cos^3 x}{x\sin 2x}$$ There is a rule that says, that the limit of difference equals the difference of limits if both the limits in the difference of limits exist. So in this case the limit $\lim_{x \to 0} \frac{1}{x\sin 2x}$ does not exist as it is not approaching any particular value. So the question is, why does the limit of difference exist if the difference of those two limits doesn't?

  • 1
    $\begingroup$ Check the limit btw, it should be zero. $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 14:11
  • 2
    $\begingroup$ @SimplyBeautifulArt: I make it $\frac34$. So does WolframAlpha. $\endgroup$ – TonyK Jan 22 '17 at 14:15
  • $\begingroup$ @TonyK Oops, my bad. I went back to my calculations and I accidentally changed $\cos^3(x)$ to $\cos(x^3)$ in a step XD $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 16:09

The separation of limits into the form

$$\lim_{x\to a}f(x)-g(x)=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)$$

only holds when the limits exist. In this case, they don't, so we simply aren't allowed to do that, else we would always have

$$\lim_{x\to a}f(x)=\lim_{x\to a}f(x)-\frac1{x-a}+\frac1{x-a}=\underbrace{\lim_{x\to a}f(x)-\frac1{x-a}+\lim_{x\to a}\frac1{x-a}}_{\text{undefined}}$$

  • 3
    $\begingroup$ I see... I was wrongly using this "method" to prove that certain limits do not exist, but in this case I was confused as the results were contradictory. So now the new question arises, what is the correct method to see that certain limit does not exist? For example $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x^2}$ $\endgroup$ – antestor Jan 22 '17 at 14:16
  • $\begingroup$ @antestor Epsilon delta? $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 16:10
  • $\begingroup$ Yes, to check the existence of limit by definition it should yield the correct answer, but it's tedious and not one of the easiest methods. I thought there are other (faster) ways. $\endgroup$ – antestor Jan 22 '17 at 16:14
  • $\begingroup$ @antestor Well, showing that the left and right side limits are unequal, setting inequalities, etc. These can show the limit doesn't exist. $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 16:16

You have the illustration that two expressions may tend to infinity, yet they approach each other. What's so extraordinary with this? It happens very commonly with asymptotes.

That say, computing the limit is much more illuminating using Taylor's polynomials: you see in depth why the limit is what it is:

First, at order $2$, we know that $\;\cos x=1-\dfrac{x^2}2+o(x^2)$, so $$1-\cos^3 x=1-\Bigl[\Bigl(1-\dfrac{x^2}2+o(x^2)\Bigr)^3\Bigr]=1-\Bigl[1-\dfrac{3x^2}2+o(x^2)\Bigr]=\dfrac{3x^2}2+o(x^2)$$ On the other hand, $\;x\sin 2x=x\bigl(2x+o(x)\bigr)=2x^2+o(x^2)$, so $$\frac{1-\cos^3 x}{x\sin 2x}=\frac{\dfrac{3x^2}2+o(x^2)}{2x^2+o(x^2)}=\frac{\dfrac32+o(1)}{2+o(1)}\to\dfrac34.$$


If the limits of two functions exist then the limit of their difference (sum, product, quotient provided limit of divisor is non-zero) also exists. This is a standard result proven in most textbooks and you are aware of this result.

From this result it does not follow that if the limits of two functions do not exist then the limit of their difference does not exist. If $A\Rightarrow B$ then it does not mean that $\neg A\Rightarrow \neg B$.

It is quite possible like in your question that limits of two functions do not exist and yet the limit of their difference exists. In this particular example in your question you may find it difficult to conclude that the limit of difference exists. So consider the simpler case when $\lim_{x\to 0}(1/x\sin x)$ does not exist and yet $$\lim_{x\to 0}\frac{1}{x\sin x} - \frac{1}{x\sin x} =0 $$ in an obvious manner. There may be similar cases where the limit of difference may not exist. Thus $$\lim_{x\to 0}\frac{1}{x\sin x} - \frac{1}{x}$$ does not exist.

If the hypotheses of a theorem do not hold then it is simply not possible to say anything in general about the conclusions of the theorem. And this is true for the limit theorems which I mentioned at the start of the answer.

The limit in question exists and its existence can be proved via its evaluation as follows \begin{align} L&=\lim_{x\to 0}\frac{1-\cos^{3}x}{x\sin 2x}\notag\\ &=\lim_{x\to 0}\frac{1-\cos^{3}x}{1-\cos x} \cdot\frac{1-\cos x} {x\cdot 2x}\cdot\frac{2x}{\sin 2x}\notag \\ &=\lim_{t\to 1}\frac{t^{3}-1}{t-1}\cdot\lim_{x\to 0}\frac{1-\cos x} {2x^{2}}\cdot 1\text{ (putting }t=\cos x) \notag\\ &=\frac{3}{2}\lim_{x\to 0}\frac{1-\cos x} {x^{2}}\notag\\ &=\frac{3}{2}\cdot\frac{1}{2}\notag\\ &=\frac{3}{4}\notag \end{align}


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.