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^{6}-53x^{3}+56}{3x^{2}+18 + 7x^{3}+42x}

Explanation:

Step1: Identify highest - degree terms

For the numerator $12x^{6}-53x^{3}+56$, the highest - degree term is $12x^{6}$. For the denominator $3x^{2}+18 + 7x^{3}+42x$, the highest - degree term is $7x^{3}$.

Step2: Divide numerator and denominator by $x^{3}$ (the highest - degree of the denominator)

\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{12x^{6}-53x^{3}+56}{3x^{2}+18 + 7x^{3}+42x}&=\lim_{x ightarrow\infty}\frac{\frac{12x^{6}}{x^{3}}-\frac{53x^{3}}{x^{3}}+\frac{56}{x^{3}}}{\frac{3x^{2}}{x^{3}}+\frac{18}{x^{3}}+\frac{7x^{3}}{x^{3}}+\frac{42x}{x^{3}}}\\ &=\lim_{x ightarrow\infty}\frac{12x^{3}-53+\frac{56}{x^{3}}}{\frac{3}{x}+\frac{18}{x^{3}} + 7+\frac{42}{x^{2}}} \end{align*}$$

\]

Step3: Evaluate the limit

As $x
ightarrow\infty$, $\frac{3}{x}
ightarrow0$, $\frac{18}{x^{3}}
ightarrow0$, $\frac{56}{x^{3}}
ightarrow0$, $\frac{42}{x^{2}}
ightarrow0$.
So, $\lim_{x
ightarrow\infty}\frac{12x^{3}-53+\frac{56}{x^{3}}}{\frac{3}{x}+\frac{18}{x^{3}} + 7+\frac{42}{x^{2}}}=\infty$.

Answer:

The limit does not exist (DNE)