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**Soroban** Hello, DannyMath!

Multiply numerator and denominator by the conjugate:

. . $\displaystyle \displaystyle \frac{\sqrt{x^2+5x+1} - x}{1}\cdot\frac{\sqrt{x^2+5x+1} + x}{\sqrt{x^2+5x+1} + x}$

. . $\displaystyle \displaystyle =\;\frac{(x^2+5x +1) - x^2}{\sqrt{x^2+5x+1} + x} \;=\; \frac{5x+1}{\sqrt{x^2+5x+1} + x}$

$\displaystyle \displaystyle \text{Divide numerator and denominator by }x\!:\;\; \displaystyle \frac{\dfrac{5x+1}{x}}{\dfrac{\sqrt{x^2+5x+1} + x}{x}}\;\;\bf[1] $

$\displaystyle \displaystyle \text{The numerator is: }\;\frac{5x}{x} + \frac{1}{x} \;=\;5 + \frac{1}{x}$

$\displaystyle \displaystyle \text{The denominator is: }\;\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}$

. . $\displaystyle \displaystyle \;=\;\frac{\sqrt{x^2+5x+1}}{\sqrt{x^2}} + 1\;=\;\sqrt{\frac{x^2+5x+1}{x^2}} + 1 $

You can't do the above since $\displaystyle x<0$...it can't go into the square root. What has to be done is:

$\displaystyle \frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}=-\frac{\sqrt{x^2+5x+1}}{-x} + \frac{x}{x}=-\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=-\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1 $...etc.

At the end the limit indeed is $\displaystyle \infty$

Tonio

. . $\displaystyle \displaystyle =\;\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1 $

$\displaystyle \displaystyle \text{Then }{\bf[1]}\text{ becomes: }\; \frac{5 + \frac{1}{x}}{\sqrt{1+\frac{5}{x} + \frac{1}{x^2}} + 1} $

$\displaystyle \displaystyle \text{Therefore: }\;\lim_{x\to-\infty}\frac{5+\frac{1}{x}}{\sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1} \;=\;\frac{5 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{5}{2}$