By division law,
$$\displaystyle \lim_{x \to \infty} \dfrac{\sqrt[3]{x^2+8}}{x+2}$$
is equivalent to
$$\dfrac {\displaystyle \lim_{x \to \infty} {\sqrt[3]{x^2+8}}{}}{\displaystyle \lim_{x \to \infty} {x+2}}$$
However, the first expression evaluates to $0$ while the second expression evaluates to $\dfrac{\infty}{\infty}$ which is indeterminate. Which of them is correct or my understanding of the division law is wrong?
$\endgroup$ 72 Answers
$\begingroup$$\infty/\infty$ is an "indeterminate form" because knowing the limits of numerator and denominator does not determine the bahvior of the limit of the quotient in that case.
As Myglasses said in a comment. The limit law for a quotient only applies when the limits of the top and bottom exist (this means "is a finite number") and the denominator is not zero. (Depending on your textbook/teacher, you can also add on some non-indeterminate cases like the quotient of the limits being $\infty/(-5)$ tells you the limit is $-\infty $, or a quotient of limits like $3/\infty$ tells you the limit.of the quotient is $0$.)
$\endgroup$ 3 $\begingroup$hint: write $x+2 = \sqrt[3]{(x+2)^3}$,and use $\dfrac{\sqrt[3]{x^2+8}}{\sqrt[3]{(x+2)^3}}= \sqrt[3]{\dfrac{x^2+8}{x^3+6x^2...}}$
$\endgroup$ 3
