1)lim x^2 x->4 2)lim x^2-1//2x^2-x-1 x->1 3)lim x^2-1//2x^2-x-1 x->бесконечности 4)lim...

Решите задачу:

$\lim_{x \to 4} x^{2} = 4^{2} =16 \\ \lim_{x \to1} \frac{ x^{2} -1}{2 x^{2} -x-1} = \frac{ 1 ^{2} -1}{2 *1^{2} -1-1} = \frac{0}{0} =\ \textgreater \ \lim_{x \to1} \frac{ x^{2} -1}{2 x^{2} -x-1}= \frac{(x-1)(x+1)}{(x-1)2x+1}$ $=\frac{x+1}{2x+1} = =\frac{1+1}{2*1+1} = \frac{2}{3} \\ \lim_{x \to \infty} \frac{ x^{2} -1}{2 x^{2} -x-1} = \frac{\infty}{\infty} =\ \textgreater \ \lim_{x \to \infty} \frac{ x^{2} -1}{2 x^{2} -x-1}= \frac{1}{2} \\ \lim_{x \to 0} \frac{ x^{2} -1}{2 x^{2} -x-1}= \frac{ -1}{-1}=1 \\ 6) \lim_{x \to0} sin \frac{x ^{4} }{2x} =sin \frac{ x^{3} }{2} =0$