We need to compute the value of given limit

\(\lim_{x\rightarrow\infty}\frac{4x^3-2x-1}{x^2-1}\)

Dividing both numerator and denominator by \(x^2\) , we get

\(\lim_{x\rightarrow\infty}\frac{\frac{4x^3-2x-1}{x^2}}{\frac{x^2-1}{x^2}}\)

\(=\lim_{x\rightarrow\infty}\frac{4x-\frac{2}{x}-\frac{1}{x^2}}{1-\frac{1}{x^2}}\)

\(=\frac{\lim_{x\rightarrow\infty}(4x-\frac{2}{x}-\frac{1}{x^2})}{\lim_{x\rightarrow\infty}(1-\frac{1}{x^2})}\)

\(=\frac{(\infty-0-0)}{(1-0)}\)

\(=\infty\)

Answer: \(\therefore\lim_{x\rightarrow\infty}\frac{4x^3-2x-1}{x^2-1}=\infty\)

\(\lim_{x\rightarrow\infty}\frac{4x^3-2x-1}{x^2-1}\)

Dividing both numerator and denominator by \(x^2\) , we get

\(\lim_{x\rightarrow\infty}\frac{\frac{4x^3-2x-1}{x^2}}{\frac{x^2-1}{x^2}}\)

\(=\lim_{x\rightarrow\infty}\frac{4x-\frac{2}{x}-\frac{1}{x^2}}{1-\frac{1}{x^2}}\)

\(=\frac{\lim_{x\rightarrow\infty}(4x-\frac{2}{x}-\frac{1}{x^2})}{\lim_{x\rightarrow\infty}(1-\frac{1}{x^2})}\)

\(=\frac{(\infty-0-0)}{(1-0)}\)

\(=\infty\)

Answer: \(\therefore\lim_{x\rightarrow\infty}\frac{4x^3-2x-1}{x^2-1}=\infty\)