Could someone tell me if my proceeding is correct?
$\lim_{x\to 0}{\dfrac{\sqrt{x}\sin{\sqrt{x}}+\log(1-x)}{x-\tan{x}}} =$ $= \lim_{x\to 0}{\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{(\sqrt{x})^3}{6}+o(x^2)\right) + \left(-x -\dfrac{x^2}{2} + o(x^2)\right)}{x-\left(x+\dfrac{x^3}3+o(x^3)\right)}} =$ $= \lim_{x\to0}{\left(-\dfrac23x^2\right)\left(\dfrac{3}{x^3}\right)} =-\infty$
Taylor expansion: $\lim\limits_{x\to 0}{\frac{\sqrt{x}\sin{\sqrt{x}}+\log(1-x)}{x-\tan{x}}}$
2
$\begingroup$
calculus
limits
taylor-expansion
-
7The 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 $x \to 0^+$. – 2012-07-15