<s>[INST] 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: $\frac{-7}{4}$
 [/INST] It's a very common error: $\sqrt{x^2}=-x$, when $x<0$.
I usually suggest the substitution $t=-1/x$, so the limit becomes
$$
\lim_{t\to0^+}\left(\sqrt{\frac{4}{t^2}-\frac{7}{t}}-\frac{2}{t}\right)
=
\lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t}
$$
which is an easy derivative:
$$
f(t)=\sqrt{4-7t}
\qquad
f'(t)=\frac{-7}{2\sqrt{4-7t}}
\qquad
f'(0)=-\frac{7}{4}
$$
or, with a Taylor expansion,
$$
\lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t}
=
\lim_{t\to0^+}\frac{2(\sqrt{1-7t/4}-1)}{t}
=
\lim_{t\to0^+}\frac{2\bigl(1-\frac{1}{2}\frac{7t}{4}+o(t)\bigr)}{t}=-\frac{7}{4}
$$
</s>