Question

14.5/33
ab
continuity
$lim_{x
ightarrowinfty}\frac{x^{4}-2x^{2}+3x - 4}{2x^{5}-3x^{3}+2x - 1}=dne$
0/4
te for the graph of the function f. which of the

Explanation:

Step1: Divide numerator and denominator by highest - power of x in denominator

Divide both the numerator $x^{4}-2x^{2}+3x - 4$ and the denominator $2x^{5}-3x^{3}+2x - 1$ by $x^{5}$.
We get $\lim_{x
ightarrow\infty}\frac{\frac{x^{4}}{x^{5}}-\frac{2x^{2}}{x^{5}}+\frac{3x}{x^{5}}-\frac{4}{x^{5}}}{\frac{2x^{5}}{x^{5}}-\frac{3x^{3}}{x^{5}}+\frac{2x}{x^{5}}-\frac{1}{x^{5}}}=\lim_{x
ightarrow\infty}\frac{\frac{1}{x}-\frac{2}{x^{3}}+\frac{3}{x^{4}}-\frac{4}{x^{5}}}{2-\frac{3}{x^{2}}+\frac{2}{x^{4}}-\frac{1}{x^{5}}}$

Step2: Evaluate the limit as x approaches infinity

As $x
ightarrow\infty$, $\frac{1}{x}
ightarrow0$, $\frac{2}{x^{3}}
ightarrow0$, $\frac{3}{x^{4}}
ightarrow0$, $\frac{4}{x^{5}}
ightarrow0$, $\frac{3}{x^{2}}
ightarrow0$, $\frac{2}{x^{4}}
ightarrow0$, $\frac{1}{x^{5}}
ightarrow0$.
So, $\lim_{x
ightarrow\infty}\frac{\frac{1}{x}-\frac{2}{x^{3}}+\frac{3}{x^{4}}-\frac{4}{x^{5}}}{2-\frac{3}{x^{2}}+\frac{2}{x^{4}}-\frac{1}{x^{5}}}=\frac{0 - 0+0 - 0}{2-0 + 0-0}=0$

Answer:

$0$