Finding limits with a radical in denominator

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Homework Statement


Evaluate: ##\lim_{x \rightarrow -\infty} {\frac{3x^3+2}{\sqrt{x^4-2}}}##

Homework Equations




The Attempt at a Solution



For limits involving fractions, it's a good idea to divide the numerator and the denominator by the highest degree x in the fraction. In doing this, we can separate the constituent pieces and evaluate them individually as the limit goes to ##a##.

Now in this problem, with exponents under radicals, I am having a small hiccup.

In the numerator, we have ##x^3##, and in the denominator we have ##x^4## under a radical which can be seen as ##x^\frac{4}{2}## which is the same as ##x^2##.
So with this reasoning, I see ##x^3## as the higher degree and divide the numerator and denominator by ##x^3## and go on to break it apart.
However, in doing this, I get an indeterminate result ##\frac{0}{0}## which I do not want. Next I tried to divide by ##x^2## or ##\sqrt{x^4}##
This gives me a better solution which tells me the limit goes to -∞.

Now my question is, why did dividing by the lesser degree provide the limit, but the higher degree did not? Or is the ##x^4## the higher degree even though it's under a radical?

Thank you for reading!
 

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  • #2
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Homework Statement


Evaluate: ##\lim_{x \rightarrow -\infty} {\frac{3x^3+2}{\sqrt{x^4-2}}}##

Homework Equations




The Attempt at a Solution



For limits involving fractions, it's a good idea to divide the numerator and the denominator by the highest degree x in the fraction. In doing this, we can separate the constituent pieces and evaluate them individually as the limit goes to ##a##.

Now in this problem, with exponents under radicals, I am having a small hiccup.

In the numerator, we have ##x^3##, and in the denominator we have ##x^4## under a radical which can be seen as ##x^\frac{4}{2}## which is the same as ##x^2##.
So with this reasoning, I see ##x^3## as the higher degree and divide the numerator and denominator by ##x^3## and go on to break it apart.
However, in doing this, I get an indeterminate result ##\frac{0}{0}## which I do not want. Next I tried to divide by ##x^2## or ##\sqrt{x^4}##
This gives me a better solution which tells me the limit goes to -∞.

Now my question is, why did dividing by the lesser degree provide the limit, but the higher degree did not? Or is the ##x^4## the higher degree even though it's under a radical?

Thank you for reading!
What usually disturbs is the denominator. Thus we deal with it first. We don't need something like ##\frac{0}{0}## - and again: please forget this immediately - because we have ##x \to \infty##, so no risk for a zero in the denominator in the region which we are interested in. We change as few as possible, so
$$
\lim_{x \to \infty} \dfrac{3x^3+2}{\sqrt{x^4-2}} = \lim_{x \to \infty} \dfrac{3x+\dfrac{2}{x^2}}{\sqrt{1-\dfrac{2}{x^4}}}
$$
and now you can relax and look what happens if ##x \to \infty##. This will also work, if e.g. the nominator is of less degree.
 
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This one is basically ## \frac{3x^3}{x^2}=3x ##. The answer is obvious, but the homework helpers aren't supposed to give the final answer.
 
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What usually disturbs is the denominator. Thus we deal with it first.
What do you mean by "disturbs"? Are you saying that it's good practice to divide out the ##x## in the denominator regardless of degree?

We don't need something like 00\frac{0}{0} - and again: please forget this immediately - because we have x→∞x \to \infty, so no risk for a zero in the denominator in the region which we are interested in
I had no intention on getting ##\frac{0}{0}##, it just came to when I tried to break things up, so instead I went back and went with the other degree of x. Once I went with ##x^2##, I got what you have there and could see that the limit went to ##\frac{-∞}{1}## = ##-∞##

This one is basically ## \frac{3x^3}{x^2}=3x ##. The answer is obvious, but the homework helpers aren't supposed to give the final answer.
I would agree, but my intuition for these things is subpar at best right now, and I want to be able to be vigorous in my explanations as to what the limit approaches. Once I get these down better, I would feel more comfortable making statements like that.
 
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What do you mean by "disturbs"? Are you saying that it's good practice to divide out the ##x## in the denominator regardless of degree?
Yes. Since we must not divide by zero, it's good practice to transform the denominator in a way, which doesn't result in a zero. To get rid of the highest ##x-##term is often a way. It doesn't cost much and depending on the result we can go on or try something else.
 
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Ok great thank you.
I've been getting some of my information from brilliant.org and it's actually really good. One thing it said was what I mentioned about dividing the numerator and denominator by the highest powered x and it has worked so far. But I got a curveball for this problem which I got out of Schaum's Calculus problems (I like to get problems from different sources so I can't guess as to which method the text wants me to solve the given problem). So I'll press on in dividing out the x in the denominator and see if I come into anything new.
 

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