# Finding limits with a radical in denominator

Gold Member

## Homework Statement

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

## 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?

fresh_42
Mentor
2021 Award

## Homework Statement

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

## 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?

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.

• opus
Homework Helper
Gold Member
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.

• opus
Gold Member
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.

fresh_42
Mentor
2021 Award
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.

• opus
Gold Member
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.