Question

Evaluate the limit as x turns to -∞

Lim [3x + √(9x²-Ix-2I)]

Answer 1

$\text{Given the limit:}\\ I = \displaystyle \lim_{x \rightarrow -\infty} \left( 3x + \sqrt{9x^2 - |x-2|} \right)\\ \text{Ratinonalising the numerator we have the following: }\\ \displaystyle I = \lim_{x \rightarrow -\infty} \frac{|x-2|}{3x - \sqrt{9x^2 - |x-2|}} \\ \text{Set } x = -x \text{ then we have that: } \\ \displaystyle I = \lim_{x \rightarrow \infty} \frac{x+2}{-3x - \sqrt{9x^2 - (x+2)}} = -\lim_{x \rightarrow -\infty} \frac{x+2}{\sqrt{9x^2 - x-2} + 3x} \\ \quad = \displaystyle - \lim_{x \rightarrow \infty} \frac{\frac xx+ \frac 2x}{\frac{\sqrt{9x^2 - x-2}}{x} + \frac{3x}{x}} = - \lim_{x \rightarrow \infty} \frac{1+ \frac 2x}{\sqrt{\frac{9x^2}{x^2} - \frac{x}{x^2} - \frac{2}{x^2}} + 3} \\ \quad = - \lim_{x \rightarrow \infty} \frac{1+ \frac 2x}{\sqrt{9 - \frac{1}{x} - \frac{2}{x^2}} + 3} \\ \quad = -\frac{1+ 0}{\sqrt{9 - 0 - 0} + 3} = -\frac{1}{3 + 3} \\ \quad = -\frac16$