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: $\frac{-7}{4}$
 A: Because, when you divide the denominator by $x$, you forgot $x$ is supposed to be negative, so that
$$x=-\sqrt{x^2}.$$
A: hint  When $x$ goes to $-\infty,$ it becomes negative .
on the other hand,
we have $$\boxed{\sqrt{x^2}=|x|}$$
the mistake you made can be corrected by $$\sqrt{(-x)^2}=-x$$.
In the denominator, factor out by $(-x)^2$.
A: In these cases, in order to avoid mistakes with the sign, we can let $y=-x\to \infty$ to obtain
$$\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}=\lim_{y\to \infty} \frac{-7y}{\sqrt {4y^2-7y}+2y}=-\frac 74$$
A: Hint: It is $$\sqrt{4x^2+7x}=\sqrt{x^2\left(4+\frac{7}{x}\right)}=-x\sqrt{4+\frac{7}{x}}$$
A: 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,
$$
\lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t}
=
\lim_{t\to0^+}\frac{2(\sqrt{1-7t/4}-1)}{t}
=
\lim_{t\to0^+}\frac{2\bigl(1-\frac{1}{2}\frac{7t}{4}+o(t)\bigr)}{t}=-\frac{7}{4}
$$
