Finding the limit of $\sqrt{4x^2+x+7}+2x$ I've been working on this problem for a while now, but I can't solve it  
$$\lim\limits_{x\to-\infty}{\sqrt{4x^2+x+7}+2x}$$
I've tried multiplying by 
$$\frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}$$
but I didn't get it.  Am I missing something really obvious?  Can someone help me with this question?
 A: $$
\begin{aligned}
\lim_{x\to-\infty}\left(\sqrt{4x^2+x+7}+2x\right) &= \lim_{x\to-\infty}\frac{4x^2+x+7-4x^2}{\sqrt{4x^2+x+7}-2x}\\
& =\lim_{x\to-\infty}\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\
&=\lim_{x\to-\infty}\frac{-1-\frac{7}{x}}{\sqrt{4+\frac{1}{x}+\frac{7}{x^2}}+2}\\
&=-\frac{1}{4}
\end{aligned}
$$
A: To find $\lim_{x \to -\infty} \frac  {x+7} {\sqrt {4x^{2}+x+7} -2x}$ divide the numerator and the denominator by $-x$. The limit is $-1/4$. 
A: Hint: Multiplying by $\ \frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}\ \left(=1\right)\ $ shows that
\begin{align}
\sqrt{4x^2+x+7}+2x&=\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\
&=\frac{-1+\frac{7}{|x|}}{\sqrt{4-\frac{1}{|x|}+ \frac{7}{x^2}}+2}\ \ \text{for }\ x<0\ .
\end{align}
Can you see what the limit of this last expression is?
A: \begin{gather*}
\lim _{x\rightarrow -\infty }\sqrt{4x^{2} +x+7} -2x\ =\ \lim _{x\rightarrow -\infty }\sqrt{4x^{2}( 1+\frac{x}{4x^{2}} +\frac{7}{4x^{2}}} -2x\ =\ \ \\
\lim _{x\rightarrow -\infty } 2x\sqrt{( 1+\frac{1}{4x} +\frac{7}{4x^{2}}} -2x\ \ Apply\ Binomial\ theorem\\
\lim _{x\rightarrow -\infty } 2x\left( 1+\frac{1}{8x} +\frac{7}{8x^{2}}\right) -2x\ =\ \frac{1}{4}\\
\end{gather*}
A: Set $x=-\dfrac1h,\implies h\to0^+$
$$\sqrt{4x^2+x+7}=\sqrt{\dfrac{4-h+7h^2}{h^2}}=\dfrac{\sqrt{4-h+7h^2}}h$$
So, we need $$\lim_{h\to0^+}\dfrac{\sqrt{4-h+7h^2}-2}h=\lim_{h\to0^+}\dfrac{4-h+7h^2-2^2}{h(\sqrt{4-h+7h^2}+2)}=\dfrac{-1}{\sqrt4+2}$$
A: $$L=\lim_{x \rightarrow \infty} \sqrt{4x^2+x+7}+2x$$
$$\implies L= \lim_{x \rightarrow -\infty} |2x| \sqrt{1+\frac{1}{4x}+\frac{7}{4x^2}}+2x$$
$$\implies L=-2x\left(1+\frac{1}{8x}+\frac{7}{8x^2}+..+O(1/x^2)\right)+2x$$
$$L=-\frac{1}{4}.$$
