Question

find lim(x→∞) (5x^4 + x^2)/(2x^4 - x^3 - 4). choose 1 answer: a 5/2 b 0 c -1/4 d the limit is unbounded

Explanation:

Step1: Divide by highest - power of x in denominator

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

Step2: Evaluate the limit of each term

As $x
ightarrow\infty$, we know that $\lim_{x
ightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$.
So, $\lim_{x
ightarrow\infty}\frac{5 + 1/x^{2}}{2-1/x - 4/x^{4}}=\frac{\lim_{x
ightarrow\infty}(5)+\lim_{x
ightarrow\infty}(1/x^{2})}{\lim_{x
ightarrow\infty}(2)-\lim_{x
ightarrow\infty}(1/x)-\lim_{x
ightarrow\infty}(4/x^{4})}$.
Since $\lim_{x
ightarrow\infty}(5) = 5$, $\lim_{x
ightarrow\infty}(1/x^{2})=0$, $\lim_{x
ightarrow\infty}(2) = 2$, $\lim_{x
ightarrow\infty}(1/x)=0$, and $\lim_{x
ightarrow\infty}(4/x^{4}) = 0$, we have $\frac{5 + 0}{2-0 - 0}=\frac{5}{2}$.

Answer:

A. $\frac{5}{2}$