Evaluate the following limits algebraically:
lim x - infinity 7x^3 + 5x/(8x^9 + x^6 + 4) ^1/3

Question

Evaluate the following limits algebraically: 
lim x -> infinity 7x^3 + 5x/(8x^9 + x^6 + 4) ^1/3

kirbykirby · Answer

$$ \lim_{x\to\infty}7x^3+\frac{5x}{(8x^9+x^6+4)^{1/3}}$$ ?

anonymous · Answer

$$\lim_{x \rightarrow \infty} \frac{ 7x ^{3} + 5x}{ (8x^9 + x^6 + 4) ^1/3}$$

kirbykirby · Answer

Oh ok

$$ \large \begin{align}\lim_{x\to\infty}\frac{7x^3+5x}{(8x^9+x^6+4)^{1/3}}&=\lim_{x\to\infty}\frac{x^3\left( 7+\frac{5}{x^2}\right)}{\left(x^9\left(8+\frac{1}{x^3}+\frac{4}{x^9} \right) \right)^{1/3}}\&=\lim_{x\to\infty}\frac{x^3\left( 7+\frac{5}{x^2}\right)}{\left( 8+\frac{1}{x^3}+\frac{4}{x^9}\right)^{1/3}x^3}\&=\lim_{x\to\infty}\frac{\left( 7+\frac{5}{x^2}\right)}{\left( 8+\frac{1}{x^3}+\frac{4}{x^9}\right)^{1/3}} \&=\frac{7}{8^{1/3}}=\frac{7}{2} \end{align}$$
The before last line simplifies because all the limits for fractions with x in the denominator go to 0 as x goes to infinity

mathmath333 · Answer

$\Large\tt \color{black}{great~~~~\ddot\smile}$

kirbykirby · Answer

$\dddot \smile$ alien smile

mathmath333 · Answer

lol

anonymous · Answer

thank you both so much!!! :)

mathmath333 · Answer

yw$\large\tt \color{black}{great\ddddot\smile}$

kirbykirby · Answer

=]