$\displaystyle\lim_{x\to-\infty}2x+\sqrt{4x^2+x}$ How do I calculate
$$\lim_{x\to-\infty}2x+\sqrt{4x^2+x}?$$
I've gotten "close" many times but I just can't figure it out, I always get stuck with zero division or something similar.
 A: Define $t:=-x$. Then, 
\begin{align*}
&\lim_{x \to -\infty}2x+\sqrt{4x^2+x}=\lim_{t \to \infty} \left( -2t+\sqrt{4t^2-t} \right)\\
&=\lim_{t \to \infty} \left( -2t+\sqrt{4t^2-t} \right)\cdot\frac{2t+\sqrt{4t^2-t}}{2t+\sqrt{4t^2-t}} =\lim_{t \to \infty}\frac{-t}{2t+\sqrt{4t^2-t}} \\
&=\lim_{t \to \infty}\frac{-1}{2+\frac{\sqrt{4t^2-t}}{t}}=\lim_{t \to \infty}\frac{-1}{2+\sqrt{\frac{4t^2-t}{t^2}}}=\lim_{t \to \infty}\frac{-1}{2+\sqrt{\frac{4t^2-t}{t^2}}}\\
&=\lim_{t \to \infty}\frac{-1}{2+\sqrt{4-(1/t)}}=\frac{-1}{2+\sqrt{4}}=-1/4.
\end{align*}
A: \begin{align}\lim_{x\to-\infty}2x+\sqrt{4x^2+x}&=\lim_{x\to-\infty}\frac{\left(2x+\sqrt{4x^2+x}\right)\left(2x-\sqrt{4x^2+x}\right)}{2x-\sqrt{4x^2+x}}\\&=\lim_{x\to-\infty}\frac{-x}{2x-\sqrt{4x^2+x}}\\&=-\lim_{x\to-\infty}\frac1{2+\sqrt{4+\frac1x}}\text{ (because $x<0$)}\\&=-\frac14.\end{align}
A: $$\begin{align*} \lim_{x \rightarrow -\infty} 2x + \sqrt{4x^2 + x} & = \lim_{x \rightarrow \infty} \sqrt{4x^2 - x} - 2x = \\ & = \lim_{x \rightarrow \infty} \frac{(\sqrt{4x^2 - x} - 2x)(\sqrt{4x^2 - x} + 2x)}{\sqrt{4x^2 - x} + 2x} = \\ & = \lim_{x \rightarrow \infty} -\frac{x}{\sqrt{4x^2 - x} + 2x} = \boxed{-\frac{1}{4}} \end{align*}$$
A: Making the substitution $u = 1/x$, we aim to find
$$\lim_{u\to 0^{\large{-}}} \left(2/u + \sqrt{4/u^2 + 1/u}\ \right).$$
When $u < 0$, we have $\sqrt{u^2} = -u$, so $\sqrt{4/u^2 + 1/u} = \sqrt{(4+u)/u^2}=-\sqrt{4 + u}/u$. Hence, we obtain
$$\lim_{u\to 0^{\large{-}}} \frac{2-\sqrt{4 + u}}{u}.$$
The numerator is $0$ at $u=0$. So let $f(u) = 2 - \sqrt{4+u}$. If we use the definition of the derivative, we see that our limit can be written as
$$\lim_{u\to 0^{\large{-}}}\frac{f(u) - f(0)}{u} = f'(u)\Big|_{u=0}$$
provided the derivative exists. Applying the standard rules of differentiation, $f'(u) = -1/(2\sqrt{4+u})$. This equals $-1/4$ at $u=0$. Hence, 
$$\lim_{x\to-\infty} \left(2x + \sqrt{4x^2 +x}\right) = -1/4.$$
A: Note that for $x\to -\infty$,  $4x^2+x=(2x+\frac14)^2-\frac{1}{16} \sim (2x+\frac14)^2.$ Hence:
$$\lim_{x\to-\infty} \left(2x+\sqrt{4x^2+x}\right)=\lim_{x\to-\infty}\left(2x-\left(2x+\frac14\right)\right)=-\frac14.$$
