A Limit Question of 0/0 Uncertainty This questions of mine, I couldn't solve whatever I did. Please do not use L'hospital as it has not been taught to us and I don't think that kind of answer will be accepted in the exam.
$$\lim\limits_{x\to 0} \frac{3x + \sin^2x}{\sin2x - x^3}$$

L'Hôpitals rule is not necessary here. Just divide the numerator and the denominator by $2x$ and exploit the fact that $$\lim\limits_{x\to 0}\frac{\sin x}{x}=1$$
Here are the steps
$$\lim\limits_{x\to 0} \frac{3x + \sin^2x}{\sin2x - x^3}$$
$$=\lim\limits_{x\to 0} \frac{\frac{3}{2} + \frac{\sin^2x}{2x}}{\frac{\sin2x}{2x} - \frac{x^2}{2}}$$
$$=\frac{\frac{3}{2} + \frac12\lim\limits_{x\to 0} \frac{\sin^2x}{x}}{\lim\limits_{x\to 0} \frac{\sin2x}{2x} -\lim\limits_{x\to 0} \frac{x^2}{2}}$$
$$=\frac{\frac{3}{2} + \frac12\left(\lim\limits_{x\to 0} \frac{\sin x}{x}\right)\left(\lim\limits_{x\to 0} \sin x\right)}{\lim\limits_{x\to 0} \frac{\sin2x}{2x} -\lim\limits_{x\to 0} \frac{x^2}{2}}$$
$$=\frac{\frac{3}{2} + \frac12\cdot 1\cdot 0}{1-0}$$
$$=\frac32$$
