Question

limit(x->0) sin x^2/x=?​

Answer 1

$\displaystyle\\\lim_{x\to0}\dfrac{\sin x^2}{x}=\\\\\lim_{x\to0}\dfrac{(\sin x^2)'}{x'}=\\\\\lim_{x\to0}\dfrac{\cos x^2 \cdot 2x}{1}=\\\\\lim_{x\to0}2x\cos x^2=\\\\2\cdot 0 \cdot \cos 0^2=0$