I'm trying to evaluate the following limit: $$\lim_{x\to0^+}\frac{\sqrt{4(\tan x-\sin x)+1} -1}{\sqrt{4x^3+1}-1}$$ It is $\frac00$ indeterminate form. using Hopital rule doesn't seem to be a good idea because we get $\frac00$ again.

Taylor series of $\sin x$ and $\tan x$ are:

$$\sin x=x-\frac{x^3}6+O(x^5)$$ $$\tan x=x-\frac{x^3}3+O(x^5)$$ So $\sqrt{4(\tan x-\sin x)+1}-1\sim\sqrt{4(-\frac{x^3}6)+1}-1$. and I think we can compute following limit instead:

$$\lim_{x\to0^+}\frac{\sqrt{-\frac23x^3+1} -1}{\sqrt{4x^3+1}-1}$$ But I don't know how to continue.

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    $\begingroup$ You can also use the Maclaurin series for $\sqrt{1+u}$ to continue to simplify. Or, you can multiply the original fraction by its algebraic conjugate $$\frac{\sqrt{4(\tan x-\sin x)+1} +1}{\sqrt{4x^3+1}+1}$$ whose limit is easy to evaluate. $\endgroup$ May 23, 2021 at 0:32
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    $\begingroup$ @GregMartin Thanks! We can multiply it by $\frac{\sqrt{4(\tan x-\sin x)+1} +1}{\sqrt{4x^3+1}+1}$ because its limit is $1$ when $x\to0^+$ am I right? $\endgroup$
    – Etemon
    May 23, 2021 at 0:40

1 Answer 1


With pure algebraic transformations $$\lim\limits_{x\to0^+}\frac{\sqrt{4(\tan x-\sin x)+1} -1}{\sqrt{4x^3+1}-1}=\\ \lim\limits_{x\to0^+}\frac{4(\tan x-\sin x)+1 -1}{4x^3+1-1}\cdot\frac{\sqrt{4x^3+1}+1}{\sqrt{4(\tan x-\sin x)+1} +1}$$ Now we need to evaluate only $$\lim\limits_{x\to0^+}\frac{1 -\cos x}{x^2}=\lim\limits_{x\to0^+}\frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{2}$$ same trick works with your question.


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