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

Question

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.

Answer 1

While $x$ is large then $4x^2-7\sim4x^2$ hence

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

Answer 2

Note that $\sqrt {4x^2} = 2|x|$ and in this case $|x|=-x$, thus $$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{3x-\sqrt{4x^2}}= \lim_{x\to-\infty} \frac{5x}{3x+2x }=1$$

Answer 3

Here's my way :

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

Answer 4

Let $x<0.$ Then

$$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty} \dfrac{5+\dfrac{9}{x}}{3+\dfrac{2}{x}+\sqrt{4-\dfrac{7}{x^2}}}=\frac{5+0}{3+0+\sqrt{4-0}}=1.$$