# Finding limits with a radical in denominator

• opus
In summary: Again, thank you for responding to my questions.In summary, when evaluating limits involving fractions, it is good practice to divide the numerator and denominator by the highest degree of x. However, in some cases, dividing by a lower degree may be necessary to avoid a zero in the denominator. It is important to carefully consider the degree of x in both the numerator and denominator before dividing.
opus
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?

opus said:

## 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
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
fresh_42 said:
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?

fresh_42 said:
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.

opus said:
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
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.

## 1. What is the process for finding limits with a radical in the denominator?

Finding limits with a radical in the denominator involves first simplifying the expression by rationalizing the denominator. This means multiplying the numerator and denominator by the conjugate of the radical, which eliminates the radical in the denominator. Then, the limit can be evaluated by plugging in the value that the variable approaches to the simplified expression.

## 2. Can the limit of a function with a radical in the denominator exist?

Yes, the limit of a function with a radical in the denominator can exist as long as the limit of the simplified expression exists. However, it is important to note that if the simplified expression has a denominator of zero, the limit does not exist.

## 3. How do I know if I need to rationalize the denominator when finding a limit?

You should try to rationalize the denominator when finding a limit if the expression has a radical in the denominator and the variable approaches a value that would make the denominator equal to zero. This is because the limit would not exist in this case, so rationalizing the denominator is necessary to find the limit.

## 4. Can I use L'Hopital's rule to find the limit of a function with a radical in the denominator?

Yes, L'Hopital's rule can be used to find the limit of a function with a radical in the denominator. This rule states that if the limit of a function can be written as the form of 0/0 or ∞/∞, then the limit can be found by taking the derivative of both the numerator and denominator and evaluating the limit again.

## 5. Are there any restrictions when using the rationalization method to find limits?

Yes, there are some restrictions when using the rationalization method to find limits. This method can only be used when the limit of the simplified expression exists, and the denominator cannot equal zero. Additionally, the limit may not exist if the expression has a radical with an odd index, such as √x^3, as the simplified expression would have a negative value under the radical.

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