# $\lim_{x\to0^+}\frac{\sqrt{4(\tan x-\sin x)+1} -1}{\sqrt{4x^3+1}-1}$

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.

• 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. Commented May 23, 2021 at 0:32
• @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? Commented May 23, 2021 at 0:40

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.