Question

Doresc rezolvarea la urmatorul exercițiu.

Answer 1

$\displaystyle \lim_{x \to 0} \frac{sinx^2-ln\Big(x^2+1\Big)}{x^2} =\lim_{x \to 0} \frac{sinx^2}{x^2} -\lim_{x \to 0} \frac{ln\Big(x^2+1\Big)}{x^2} =\\\\=\lim_{x \to 0} \frac{\Big(sinx^2\Big)'}{\Big(x^2\Big)'} -\lim_{x \to 0} \frac{\Big(ln\Big(x^2+1\Big)\Big)'}{\Big(x^2\Big)'} =\\\\=\lim_{x \to 0} \frac{cosx^2 \cdot \Big(x^2\Big)'}{2x} -\lim_{x \to 0}\Bigg(\frac{\cfrac{1}{x^2+1} }{2x} \cdot \Big(x^2+1\Big)'\Bigg)=$

$\displaystyle =\lim_{x \to 0} \frac{cosx^2 \cdot 2x}{2x} -\lim_{x \to 0} \Bigg(\frac{2x}{x^2+1} \cdot \frac{1}{2x}\Bigg)=\lim_{x \to 0} cosx^2-\lim_{x \to 0} \frac{1}{x^2+1} =\\\\=cos0^2-\frac{1}{0^2+1} =cos0-\frac{1}{1} =1-1=0$