Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$? 
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$

My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.

Here's my way :
$$ \lim_{x\to -\infty}  \frac {5x+9}{3x+2-\sqrt{4x^2-7}} = \lim_{x\to -\infty} \frac {x (5+ \frac {9}{x})}{3x+2-\sqrt{4x^2(1-\frac{7}{4x^2})}}=\lim_{x\to -\infty}\frac {x (5+ \frac {9}{x})}{3x+2-2|x|\sqrt{1-\frac{7}{4x^2}}}=\lim_{x\to -\infty} \frac {x (5+ \frac {9}{x})}{3x+2+2x\sqrt{1-\frac{7}{4x^2}}} = \lim_{x\to -\infty} \frac {x (5 + \frac {9}{x})}{x (3+\frac{2}{x} +2\sqrt{1-\frac{7}{4x^2}})} =  \lim_{x\to -\infty} \frac {5+\frac{9}{x}}{3+\frac{2}{x} +2\sqrt{1-\frac{7}{4x^2}}} =  \frac {5}{5} = 1$$
