Вычислить предел lim при x0=2 3x^2-x-10 / 7x-x^2-10

Решите задачу:

$\displaystyle \lim_{x \to 2} \frac{3x^2-x-10}{7x-x^2-10}=\bigg[\frac{0}0\bigg]\\\\\\3x^2-x-10\\\\D=1+4\cdot3\cdot10=1+120=121=11^2\\\\x_{12}=\frac{1\pm11}{6}=2,\,\, -\frac{5}3\\\\3x^2-x-10 \quad \rightarrow \quad (x-2)(3x+5)\\\\\\7x-x^2-10=-x^2+7x-10=-(x^2-7x+10)\\\\D=49-40=9=3^2\\\\x_{12}=\frac{7\pm3}{2}=5,\,\,2\\\\-(x^2-7x+10)\quad \rightarrow \quad -(x-2)(x-5)\\\\\\\lim_{x \to 2} \frac{3x^2-x-10}{7x-x^2-10}=\lim_{x \to 2} \frac{(x-2)(3x+5)}{-(x-2)(x-5)}=-\lim_{x \to 2} \frac{3x+5}{x-5}=$

$=\displaystyle -\frac{3\cdot 2+5}{2-5}=-\frac{6+5}{-3}=\boxed{\frac{11}3}$