I need to evaluate the limit without using l'Hopital's rule.

$$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$

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    $\begingroup$ Are you allowed to use Taylor series? $\endgroup$ – JimmyK4542 Jun 15 '14 at 21:27
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    $\begingroup$ Well I would have to use Taylor Series as the last alternative, even l'Hopital. $\endgroup$ – user122673 Jun 15 '14 at 21:30
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    $\begingroup$ You could multiply the numerator and denominator by $\sqrt{1+x\sin x}+\sqrt{\cos 2x}$ (the conjugate of the numerator). Then you are left with a bit of a mess, but it is doable with basic algebraic manipulations. Using Taylor series is faster, assuming you know how to use them. $\endgroup$ – JimmyK4542 Jun 15 '14 at 21:35


Letting $x$ go to $0$ in the last formulation of the limit gives:



If we are allowed to use Taylor series:

Since $\sin x = x + O(x^3)$, we have $1+x\sin x = 1 + x^2 + O(x^4)$.

Thus, $\sqrt{1+x\sin x} = 1+\dfrac{1}{2}x^2 + O(x^4)$.

Since $\cos x = 1 - \dfrac{1}{2}x^2 + O(x^4)$, we have $\cos 2x = 1 - 2x^2 + O(x^4)$.

Thus, $\sqrt{\cos 2x} = 1-x^2 + O(x^4)$.

Since $\tan x = x + O(x^3)$, we have $\tan \dfrac{x}{2} = \dfrac{1}{2}x + O(x^3)$.

Thus, $\tan^2 \dfrac{x}{2} = \dfrac{1}{4}x^2 + O(x^4)$.

Therefore, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} = \dfrac{\left(1+\tfrac{1}{2}x^2 + O(x^4)\right) - \left(1-x^2+O(x^4)\right)}{\tfrac{1}{4}x^2 + O(x^4)} = \dfrac{\tfrac{3}{2}x^2+O(x^4)}{\tfrac{1}{4}x^2+O(x^4)} = \dfrac{\tfrac{3}{2}+O(x^2)}{\tfrac{1}{4}+O(x^2)}$.

Hence, as $x \to 0$, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} \to \dfrac{\tfrac{3}{2}}{\tfrac{1}{4}} = 6$.


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