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I am having trouble with a small thing in a limit. I encountered the limit $$\lim_{x\to0}\frac{\sin\sin x- x\sqrt[3]{1-x^2}}{x^5}$$ which can be solved rather easily by using Taylor expansion and calculating the quotient. The problem, though, arised when I tried the more natural approach - I tried to simplify the expression. When doing that, I simplified the fraction by separating it into the two terms, first of which is $\frac{\sin\sin x}{x^5}$. Then, I multiplied this term by $\frac{\sin x}{\sin x}$ which creates $\frac{\sin\sin x}{\sin x}×\frac{\sin x}{x^5}$. When trying to find the original limit with this expression, it is still the same as the original limit should be. Although, I noticed that when I continue by firstly taking $\lim_{x\to0}\frac{\sin\sin x}{\sin x}=1$, the overall limit then simplifies to $$\lim_{x\to0}\frac{\sin x}{x^5}-\frac{\sqrt[3]{1-x^2}}{x^4}$$ This limit, though, is equal to $\infty$, which is contrary to the original result, which should be $\frac{19}{90}$ (by using Taylor). Can anyone tell me what creates this error? I was thinking maybe the multiplying of $\frac{\sin x}{\sin x}$, but since that is just $1$ (no limits taken there), it should not pose a problem. The problem just seems to arise when I use the rule $\lim f(x)g(x)= \lim f(x)\lim g(x)$ on the first term there. Is this rule invalid in this case? If so, why? What am I missing? Sorry if this is really trivial, I am pretty tired by now and am probably missing something stupidly simple, but I really don't see it.

In advance, thanks for your time.

Caleb Stanford
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xDeeZee
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1 Answers1

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The problem is that you implicitly also applied the sum-of-limits rule: $$ \lim_{x\to0}\frac{\sin\sin x- x\sqrt[3]{1-x^2}}{x^5} = \lim_{x \to 0} \left( \frac{\sin\sin x}{x^5} \right) - \lim_{x \to 0} \left( \frac{x \sqrt[3]{1 - x^2}}{x^5} \right). $$

However, this rule does not work because the two limits do not exist as real numbers. (Or if you prefer: $\infty - \infty$ is an indeterminate form.)

It's necessary to apply this rule first before doing your trick of multiplying by $\frac{\sin x}{\sin x}$. Otherwise, you can't "distribute" the $\frac{\sin x}{\sin x}$ into the first term only -- you would have to multiply the second term by $\frac{\sin x}{\sin x}$ as well.

For a simpler example of what went wrong, consider $$ \lim_{x \to \infty} (x - x) = 0. $$ We can write this as $$ \lim_{x \to \infty} \left( \frac{x+1}{x+1} \cdot x - x\right) = \lim_{x \to \infty} \left( \frac{x}{x+1} \cdot (x+1) - x\right), $$ then according to the logic you used, since $\frac{x}{x+1} \to 1$, we can simplify this to $$ \lim_{x \to \infty} \left( (x + 1) - x\right) = 1. $$

Essentially, the product-of-limit rules does not apply if you are only applying it to some subterm of the limit. For a limit of the form $\lim (A + B)$, you can't apply the product-of-limit rule to just $A$ on its own.

Caleb Stanford
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  • thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something? – xDeeZee Mar 13 '19 at 18:25
  • @xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer. – Caleb Stanford Mar 13 '19 at 18:34
  • All good, then. I understand the problem. – xDeeZee Mar 13 '19 at 18:39