# Taylor expansion: $\lim\limits_{x\to 0}{\frac{\sqrt{x}\sin{\sqrt{x}}+\log(1-x)}{x-\tan{x}}}$

Could someone tell me if my proceeding is correct?
$$\lim_{x\to 0}{\dfrac{\sqrt{x}\sin{\sqrt{x}}+\log(1-x)}{x-\tan{x}}} =$$ $$= \lim_{x\to 0}{\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{(\sqrt{x})^3}{6}+o(x^2)\right) + \left(-x -\dfrac{x^2}{2} + o(x^2)\right)}{x-\left(x+\dfrac{x^3}3+o(x^3)\right)}} =$$ $$= \lim_{x\to0}{\left(-\dfrac23x^2\right)\left(\dfrac{3}{x^3}\right)} =-\infty$$

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Looks correct. Only notice that the little-o for the sine should be $o(x^4)$. – bartgol Jul 15 '12 at 16:24
The denominator should be $-3/x^3$ not $3/x^3$. So the limit is $+\infty$. Otherwise, it's correct. (I assume that the function is defined on $(0,1)$.) – Yury Jul 15 '12 at 16:52
The limit doesn't exist since $x \to 0^-$ doesn't make sense due to the presence of $\sqrt{x}$, unless by $x \to 0$, you mean $x \to 0^+$. – user17762 Jul 15 '12 at 17:50