Given the limit: I = lim x → − ∞ ( 3 x + 9 x 2 − ∣ x − 2 ∣ ) Ratinonalising the numerator we have the following: I = lim x → − ∞ ∣ x − 2 ∣ 3 x − 9 x 2 − ∣ x − 2 ∣ Set x = − x then we have that: I = lim x → ∞ x + 2 − 3 x − 9 x 2 − ( x + 2 ) = − lim x → − ∞ x + 2 9 x 2 − x − 2 + 3 x = − lim x → ∞ x x + 2 x 9 x 2 − x − 2 x + 3 x x = − lim x → ∞ 1 + 2 x 9 x 2 x 2 − x x 2 − 2 x 2 + 3 = − lim x → ∞ 1 + 2 x 9 − 1 x − 2 x 2 + 3 = − 1 + 0 9 − 0 − 0 + 3 = − 1 3 + 3 = − 1 6 \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 Given the limit: I = x → − ∞ lim ( 3 x + 9 x 2 − ∣ x − 2∣ ) Ratinonalising the numerator we have the following: I = x → − ∞ lim 3 x − 9 x 2 − ∣ x − 2∣ ∣ x − 2∣ Set x = − x then we have that: I = x → ∞ lim − 3 x − 9 x 2 − ( x + 2 ) x + 2 = − x → − ∞ lim 9 x 2 − x − 2 + 3 x x + 2 = − x → ∞ lim x 9 x 2 − x − 2 + x 3 x x x + x 2 = − x → ∞ lim x 2 9 x 2 − x 2 x − x 2 2 + 3 1 + x 2 = − x → ∞ lim 9 − x 1 − x 2 2 + 3 1 + x 2 = − 9 − 0 − 0 + 3 1 + 0 = − 3 + 3 1 = − 6 1
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