evaluate \\( \\lim _ { x \\rightarrow \\infty } \\frac { - 6 x ^ { 2 } + 3 x + 4 } { - 2 x ^ { 3 } - 4 x ^ {…

evaluate \\( \\lim _ { x \\rightarrow \\infty } \\frac { - 6 x ^ { 2 } + 3 x + 4 } { - 2 x ^ { 3 } - 4 x ^ { 2 } + 6 x - 3 } = \\)

evaluate \\( \\lim _ { x \\rightarrow \\infty } \\frac { - 6 x ^ { 2 } + 3 x + 4 } { - 2 x ^ { 3 } - 4 x ^ { 2 } + 6 x - 3 } = \\)

Answer

Explanation:

Step1: Divide numerator and denominator by (x^3)

$$\lim_{x\rightarrow\infty}\frac{\frac{-6x^{2}}{x^{3}}+\frac{3x}{x^{3}}+\frac{4}{x^{3}}}{\frac{-2x^{3}}{x^{3}}-\frac{4x^{2}}{x^{3}}+\frac{6x}{x^{3}}-\frac{3}{x^{3}}}$$ Simplify to get $$\lim_{x\rightarrow\infty}\frac{\frac{-6}{x}+\frac{3}{x^{2}}+\frac{4}{x^{3}}}{-2-\frac{4}{x}+\frac{6}{x^{2}}-\frac{3}{x^{3}}}$$

Step2: Apply the limit

As (x\rightarrow\infty), (\frac{1}{x}\rightarrow0), (\frac{1}{x^{2}}\rightarrow0), (\frac{1}{x^{3}}\rightarrow0) Substitute these values into the expression: $$\frac{0 + 0+0}{-2-0 + 0-0}$$

Answer:

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