Question

lim_{x\to\infty}\frac{x^{3}-3x^{2}+3x - 4}{4x^{3}-3x^{2}+2x - 1}=
a 4
b 1
c 1/4
d 0

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 3.
$\lim_{x
ightarrow\infty}\frac{x^{3}-3x^{2}+3x - 4}{4x^{3}-3x^{2}+2x - 1}=\lim_{x
ightarrow\infty}\frac{\frac{x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}+\frac{3x}{x^{3}}-\frac{4}{x^{3}}}{\frac{4x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}+\frac{2x}{x^{3}}-\frac{1}{x^{3}}}$

Step2: Simplify the expression

$\lim_{x
ightarrow\infty}\frac{1-\frac{3}{x}+\frac{3}{x^{2}}-\frac{4}{x^{3}}}{4-\frac{3}{x}+\frac{2}{x^{2}}-\frac{1}{x^{3}}}$
As $x
ightarrow\infty$, $\frac{1}{x}
ightarrow0$, $\frac{1}{x^{2}}
ightarrow0$, and $\frac{1}{x^{3}}
ightarrow0$.

Step3: Evaluate the limit

Substitute the values of the terms with $x$ in the denominator approaching 0.
$\frac{1 - 0+0 - 0}{4-0 + 0-0}=\frac{1}{4}$

Answer:

C. $\frac{1}{4}$