Finding limits with a radical in denominator

[opus]

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!

 

[fresh_42]

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.

 

[Charles Link]

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]

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]

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.