lim_{x\\to\\infty}\\frac{x^{3}-2x^{2}+3x - 4}{4x^{3}-3x^{2}+2x - 1}=\na 4\nb 1\nc 1/4\nd 0\ne -1

Explanation:

Step1: Divide by highest - power of x

Divide both the numerator and denominator by $x^{3}$, since the highest - power of $x$ in the denominator is $x^{3}$. $\lim_{x\rightarrow\infty}\frac{x^{3}-2x^{2}+3x - 4}{4x^{3}-3x^{2}+2x - 1}=\lim_{x\rightarrow\infty}\frac{\frac{x^{3}}{x^{3}}-\frac{2x^{2}}{x^{3}}+\frac{3x}{x^{3}}-\frac{4}{x^{3}}}{\frac{4x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}+\frac{2x}{x^{3}}-\frac{1}{x^{3}}}$

Step2: Simplify the expression

Simplify each term: $\lim_{x\rightarrow\infty}\frac{1-\frac{2}{x}+\frac{3}{x^{2}}-\frac{4}{x^{3}}}{4-\frac{3}{x}+\frac{2}{x^{2}}-\frac{1}{x^{3}}}$

Step3: Apply limit rules

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{3}}=0$. So, $\frac{\lim_{x\rightarrow\infty}(1)-2\lim_{x\rightarrow\infty}(\frac{1}{x})+3\lim_{x\rightarrow\infty}(\frac{1}{x^{2}})-4\lim_{x\rightarrow\infty}(\frac{1}{x^{3}})}{\lim_{x\rightarrow\infty}(4)-3\lim_{x\rightarrow\infty}(\frac{1}{x})+2\lim_{x\rightarrow\infty}(\frac{1}{x^{2}})-\lim_{x\rightarrow\infty}(\frac{1}{x^{3}})}=\frac{1 - 0+0 - 0}{4-0 + 0-0}=\frac{1}{4}$

Answer:

C. $1/4$