Question

find $lim_{x
ightarrowinfty}\frac{4x^{2}-5x^{3}+5}{6x^{3}+20}$ if it exists (enter na if limit does not exist, express fractions out to 3 decimal places)

Explanation:

Step1: Divide by highest - power of x

When finding the limit as $x\to\infty$ of a rational function $\frac{4x^{2}-5x^{3}+5}{6x^{3}+20}$, divide both the numerator and denominator by $x^{3}$, the highest - power of $x$ in the denominator.
We get $\lim_{x\to\infty}\frac{\frac{4x^{2}}{x^{3}}-\frac{5x^{3}}{x^{3}}+\frac{5}{x^{3}}}{\frac{6x^{3}}{x^{3}}+\frac{20}{x^{3}}}=\lim_{x\to\infty}\frac{\frac{4}{x}-5 + \frac{5}{x^{3}}}{6+\frac{20}{x^{3}}}$.

Step2: Evaluate limits of individual terms

We know that $\lim_{x\to\infty}\frac{a}{x^{n}} = 0$ for $a$ being a constant and $n>0$.
So, $\lim_{x\to\infty}\frac{4}{x}=0$, $\lim_{x\to\infty}\frac{5}{x^{3}} = 0$, and $\lim_{x\to\infty}\frac{20}{x^{3}}=0$.
Then, $\lim_{x\to\infty}\frac{\frac{4}{x}-5+\frac{5}{x^{3}}}{6+\frac{20}{x^{3}}}=\frac{0 - 5+0}{6 + 0}$.

Step3: Simplify the result

$\frac{-5}{6}\approx - 0.833$.

Answer:

$-0.833$