Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\sqrt {4x^2+7x}-2x}$$
Dividing numerator and denominator by x:
$$=\lim_{x\to -\infty} \frac{7}{\sqrt{4+\frac{7}{x}}-2}$$
$$= \frac{7}{\sqrt{4+\frac{7}{-\infty}}-2}$$
$$= \frac{7}{\sqrt{4+0}-2}$$
$$=\frac{7}{2-2}$$
$$=\infty$$
Conclusion: Limit does not exist.
Why is my solution wrong?
Correct answer: $\frac{-7}{4}$

In these cases, in order to avoid mistakes with the sign, we can let $y=-x\to \infty$ to obtain
$$\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}=\lim_{y\to \infty} \frac{-7y}{\sqrt {4y^2-7y}+2y}=-\frac 74$$
