determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
lim_{x
ightarrowinfty}\frac{12x^{2}-54x^{5}-10 + 45x^{3}}{7x^{5}+56x^{3}+2x^{2}+16}

Question

determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
 lim_{x
ightarrowinfty}\frac{12x^{2}-54x^{5}-10 + 45x^{3}}{7x^{5}+56x^{3}+2x^{2}+16}

Accepted Answer

# Explanation:
## Step1: Divide by highest - power of x
When finding the limit as $x\to\infty$ of a rational function, divide both the numerator and denominator by the highest - power of $x$ in the denominator, which is $x^{5}$.
$$
\begin{align*}
\lim_{x\to\infty}\frac{12x^{2}-54x^{5}-10 + 45x^{3}}{7x^{5}+56x^{3}+2x^{2}+16}&=\lim_{x\to\infty}\frac{\frac{12x^{2}}{x^{5}}-\frac{54x^{5}}{x^{5}}-\frac{10}{x^{5}}+\frac{45x^{3}}{x^{5}}}{\frac{7x^{5}}{x^{5}}+\frac{56x^{3}}{x^{5}}+\frac{2x^{2}}{x^{5}}+\frac{16}{x^{5}}}\
&=\lim_{x\to\infty}\frac{\frac{12}{x^{3}}-54-\frac{10}{x^{5}}+\frac{45}{x^{2}}}{7+\frac{56}{x^{2}}+\frac{2}{x^{3}}+\frac{16}{x^{5}}}
\end{align*}
$$
## Step2: Use limit rules for $\frac{1}{x^{n}}$ as $x\to\infty$
We know that $\lim_{x\to\infty}\frac{1}{x^{n}} = 0$ for $n>0$.
$$
\begin{align*}
\lim_{x\to\infty}\frac{\frac{12}{x^{3}}-54-\frac{10}{x^{5}}+\frac{45}{x^{2}}}{7+\frac{56}{x^{2}}+\frac{2}{x^{3}}+\frac{16}{x^{5}}}&=\frac{\lim_{x\to\infty}\frac{12}{x^{3}}-\lim_{x\to\infty}54-\lim_{x\to\infty}\frac{10}{x^{5}}+\lim_{x\to\infty}\frac{45}{x^{2}}}{\lim_{x\to\infty}7+\lim_{x\to\infty}\frac{56}{x^{2}}+\lim_{x\to\infty}\frac{2}{x^{3}}+\lim_{x\to\infty}\frac{16}{x^{5}}}\
&=\frac{0 - 54-0 + 0}{7+0+0+0}
\end{align*}
$$
## Step3: Simplify the result
$$
\frac{0 - 54-0 + 0}{7+0+0+0}=-\frac{54}{7}
$$
# Answer:
$-\frac{54}{7}$

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QUESTION IMAGE

Question

Explanation:

Step1: Divide by highest - power of x

Step2: Use limit rules for $\frac{1}{x^{n}}$ as $x\to\infty$

Step3: Simplify the result

Answer: