Can we make it $\frac{0}{0}$ ?

Relevant wiki: Limits by conjugates

$\begin{aligned} L & = \lim_{x \to - \infty} \left(2x + \sqrt{4x^2+3x-2} \right) \\ & = \lim_{x \to - \infty} \left(2x + \sqrt{4x^2+3x-2} \right) \times \frac {\sqrt{4x^2+3x-2}-2x}{\sqrt{4x^2+3x-2}-2x} \\ & = \lim_{x \to - \infty} \frac {4x^2+3x-2-4x^2}{\sqrt{4x^2+3x-2}-2x} \\ & = \lim_{x \to - \infty} \frac {3x-2}{\sqrt{4x^2+3x-2}-2x} & \small \color{#3D99F6} \text{Let }u = -x \\ & = \lim_{u \to \infty} \frac {-3u-2}{\sqrt{4u^2-3u-2}+2u} \\ & = \lim_{u \to \infty} \left(\frac {-3u}{\sqrt{4u^2-3u-2}+2u} - \frac 2{\sqrt{4u^2-3u-2}+2u} \right) & \small \color{#3D99F6} \text{Divide first term up and down by }u \\ & = \lim_{u \to \infty} \left(\frac {-3}{\sqrt{4-\frac 3u-\frac 2{u^2}}+2} - 0 \right) \\ & = - \frac 34 = \boxed{-0.75} \end{aligned}$

Relevant wiki: Limits by Rationalization

Let $x = -t$

$\begin{aligned} L & = \displaystyle \lim_{t \to \infty} \left( -2t +2t\sqrt{1-\dfrac{3}{4t}-\dfrac{1}{2t^2} } \right) \\ & = \displaystyle \lim_{t \to \infty} \left( -\cancel{2t} + 2t\left(\cancel{1}-\dfrac{3}{8t}-\dfrac{1}{4t^2} \right) \right) \end{aligned}$

$\boxed{L = -0.75}$

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$\begin{aligned} L & = \displaystyle \lim_{t \to \infty} \dfrac{4t^2-3t-2-4t^2}{\sqrt{4t^2-3t-2}+2t} \\ & = \displaystyle \lim_{t \to \infty} \dfrac{-3-\dfrac{2}{t}}{2\sqrt{1-\dfrac{3}{4t}-\dfrac{1}{2t^2}}+2} \end{aligned}$

$\boxed{L = -0.75}$

Take 4x^2 common from square root and take square root as -2x as x is negative and then use binomial approximation to get

2x-2x(1+3/8x) = -3/4