Вычислите пределы, пожалуйста.

Решите задачу:

$4x^2+13x+3 = 0\\ D = 169-4*4*3 = 169 - 48 = 121\\ x_1 = \frac{-13+11}{8}=- \frac{1}{4} \\ x_1 = \frac{-13-11}{8}=- 3 \\ 4x^2+13x+3 = 4(x+\frac{1}{4})(x+3)\\ ====================\\ x^2+6x+9 = (x+3)^2\\ \lim_{x \to \ -3}} \frac{4x^2+13x+3}{ x^2+6x+9} = \lim_{x \to \ -3} \frac{4(x+\frac{1}{4})(x+3)}{ (x+3)^2}= \\ \lim_{x \to \ -3} \frac{4(x+\frac{1}{4})}{ x+3}=\infty$