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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?

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    Is $x^3$ inside the cube root? – 2017-01-22
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    $\lim\dfrac{f}{g}=\dfrac{\lim f}{\lim g}$ ic correct when ${\lim f}$ and ${\lim g}$ are exists. – 2017-01-22
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    @DeepSea Oh, it's a $x^2$ actually. Yes it's inside the cube root. – 2017-01-22
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    @MyGlasses Yea that's what I understand from the law. But in the second expression limits for both numerator and denominator exists. – 2017-01-22
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    How are they.?. – 2017-01-22
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    @Tigerhix, "the limit exists" has a technical meaning that excludes $\pm\infty $. – 2017-01-22
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    @MyGlasses Ahh yes. Thanks... I now realize approaching to infinity means no limit exist... – 2017-01-22

2 Answers 2

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$\infty/\infty$ 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 $0$.)

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    Thank you very much! The textbook we use has practice problems that stated "Evaluate the following limits" and of many of them the answer is infinity. That lead me to think that somehow approaching infinity is also a kind of limit. Should've checked on the definition before asking. Thanks again, finally learnt the meaning of "indeterminate form". :) – 2017-01-22
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    @Tigerhix, It's tricky because "the limit equals $\infty$" may be a way to read an equation in a textbook even if it's also the case that "the limit doesn't exist". – 2017-01-22
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    If the downvoter would like to comment or ask a question either on Stackexchange or privately, I'd be happy to hear feedback/address something. – 2017-01-22
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hint: write $x+2 = \sqrt[3]{(x+2)^3}$,and use $\dfrac{\sqrt[3]{x^2+8}}{\sqrt[3]{(x+2)^3}}= \sqrt[3]{\dfrac{x^2+8}{x^3+6x^2...}}$

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    My method is similar which is cubing both numerator and denominator, and adding a ^(1/3) at the outermost. Which eventually evaluates to zero, same as what your steps would lead to. However I don't understand why when I use the division law, I would come to an indeterminate answer (which is infinity divided by infinity) :/ – 2017-01-22
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    Thanks anyway for the help! – 2017-01-22
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    You don't appears to be addressing TigerHix's question, just giving a hint as to why the limit of the quotient is zero (which they mentioned in the question). – 2017-01-22