Can L'hopital save you?

Substitute y = -1/x and use L'Hopital.

$\lim_{x\to -\infty} \sqrt{25x^{2}-3x}+5x$ $=\lim_{y\to 0^{+}}\frac{\sqrt{3y+25}-5}{y}$ $=\lim_{y\to 0^{+}} \frac{3}{2\sqrt{3y+25}}=\frac{3}{10}$

When we put the value of $x$ directly in the equation we get the limit in an indeterminate form like this : $\infty-\infty$

If we put $y = -x$ ,such that when $x \to -\infty$ , $y \to \infty$ , we get the limit as, $\lim_{y \to \infty} \left(\sqrt{25y^{2}+3y} -5y \right)$ Now we can rationalize the limit by multiplying the denominator and the numerator with $\sqrt{25y^{2}+3y} +5y$ , $\lim_{y \to \infty} \frac{25y^{2}+3y -25y^{2}}{\sqrt{25y^{2}+3y} +5y}$ $\Rightarrow \lim_{y \to \infty} \frac{3y}{\sqrt{25y^{2}+3y} +5y}$ We know that, $\lim_{x \to \infty} \frac{1}{x^{n}} = 0 , \text{where n= 1,2,3......}$ Using the above concept we can take out $y$ from the numerator and denominator and we get the below limit as the final answer, $\lim_{y \to \infty} \frac{3}{\sqrt{25 +\frac{3}{y}}+5} \Rightarrow \frac{3}{5+5} \Rightarrow \boxed {\frac{3}{10}}$

PS:I did notice that for $\displaystyle \lim_{x \rightarrow -\infty} \left(\sqrt{(px)^{2} + qx}+px \right)$ , the limit can be calculated using the following formula , $L = \frac{q}{2p}$