Can a limit of a real valued function $f(x)$ like $\lim_{x \to 0}\frac{f(x)}{x^2\sin(x)}$ be written as $\lim_{x \to 0}\frac{f(x)}{x^3}$?Under what circumstances is the conversion valid?

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    $\begingroup$ If any of the limits exists they are equal due to $\sin \sim_0\text{id}$ $\endgroup$
    – Git Gud
    Sep 22, 2015 at 13:56
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    $\begingroup$ Why not prove this yourself? You would probably learn a lot more than having Austin do it for you. $\endgroup$
    – GEdgar
    Sep 22, 2015 at 14:39
  • $\begingroup$ @GEdgar I did prove it myself at the first go itself...but I felt that it sometimes makes limits problems simpler than expected...so I just wanted to ensure there is no loophole in my logic .Thanks for the advice. $\endgroup$
    – user220382
    Sep 22, 2015 at 14:46
  • $\begingroup$ It is perfectly valid without any constraints. But you need to use the fact that $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. Skipping this small step is not a good idea however. $\endgroup$
    – Paramanand Singh
    Oct 4, 2015 at 5:30

1 Answer 1


You can infer the answer that you want by using the product rule for limits. Write $$\lim_{x\to 0}\frac{f(x)}{x^2 \sin(x)} = \lim_{x\to 0} \left( \frac{f(x)}{x^3} \frac{x}{\sin(x)} \right)$$ If the limit of $\frac{f(x)}{x^3}$ exists, then you can apply the product rule to get $$\lim_{x\to 0} \left( \frac{f(x)}{x^3} \frac{x}{\sin(x)} \right) = \left( \lim_{x\to 0} \frac{f(x)}{x^3} \right) \left( \lim_{x\to 0} \frac{x}{\sin(x)} \right) = \left( \lim_{x\to 0} \frac{f(x)}{x^3} \right) \cdot 1$$

  • $\begingroup$ existence of limit of $f(x)/x^3$ is not needed here. $\endgroup$
    – Paramanand Singh
    Oct 4, 2015 at 5:31

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