Question

find the limit of the rational function (a) as x→∞ and (b) as x→ - ∞. write ∞ or - ∞ where appropriate. f(x)=\frac{6x^{4}+4}{x^{4}-x^{2}+x + 7} a. lim_{x
ightarrowinfty}(\frac{6x^{4}+4}{x^{4}-x^{2}+x + 7})=square (simplify your answer.)

Explanation:

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

Divide both the numerator and denominator of $\frac{6x^{4}+4}{x^{4}-x^{2}+x + 7}$ by $x^{4}$. We get $\lim_{x
ightarrow\infty}\frac{6+\frac{4}{x^{4}}}{1-\frac{1}{x^{2}}+\frac{1}{x^{3}}+\frac{7}{x^{4}}}$.

Step2: Use limit rules

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$. So, $\lim_{x
ightarrow\infty}\frac{4}{x^{4}}=0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{3}}=0$ and $\lim_{x
ightarrow\infty}\frac{7}{x^{4}} = 0$.

Step3: Calculate the limit

Substitute these values into the function: $\frac{6 + 0}{1-0 + 0+0}=6$.

Answer:

6