# Limit problem

## Homework Statement

Find $$lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}$$

N/A

## The Attempt at a Solution

Factoring out $$\frac {(-x^6)^{1/3}}{-x^2}$$ leaves me with $$\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2 }}$$ Taking the limit at infinity gives me $$\frac{-1}{-4+3^{1/2}}$$, which is wrong.

## Answers and Replies

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Quantum Defect
Homework Helper
Gold Member

## Homework Statement

Find $$lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}$$

N/A

## The Attempt at a Solution

Factoring out $$\frac {(-x^6)^{1/3}}{-x^2}$$ leaves me with $$\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2 }}$$ Taking the limit at infinity gives me $$\frac{-1}{-4+3^{1/2}}$$, which is wrong.
Why not factor out (x^2/x^2) on the top?... i.e. lim x-> -inf [1+8/x^6]^(1/3) / [4 + SQRT(3+ 1/x^4)]

Svein
Your arithmetic is wrong (in the denominator).

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find $$lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}$$

N/A

## The Attempt at a Solution

Factoring out $$\frac {(-x^6)^{1/3}}{-x^2}$$ leaves me with $$\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2 }}$$ Taking the limit at infinity gives me $$\frac{-1}{-4+3^{1/2}}$$, which is wrong.
Why not use $(x^6+8)^{1/3} = (x^6)^{1/3} (1 + 8 x^{-6})^{1/3} \doteq x^2 [1 + (8/3)x^{-6} + O(x^{-12}) ]$, and a similar result for $(3 x^4 + 1)^{1/2}$?

Your arithmetic is wrong (in the denominator).
Ah, you're right. it should be MINUS root 3 because $$\sqrt{x^4}=-x^2$$

\doteq x^2 [1 + (8/3)x^{-6} + O(x^{-12}) ]
I don't understand how you arrived at this part.

Why not factor out (x^2/x^2) on the top?... i.e. lim x-> -inf [1+8/x^6]^(1/3) / [4 + SQRT(3+ 1/x^4)]
If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?

Quantum Defect
Homework Helper
Gold Member
[snip]

If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?
Factoring out +x^2 says nothing about what the sign of x is, does it?

The function whose limit is being evaluated is an even function of x, no? f(-x) = f(x) Wouldn't they be expected to have the same limit for very large positive and very large negative values of x?

Ray Vickson
Homework Helper
Dearly Missed
Ah, you're right. it should be MINUS root 3 because $$\sqrt{x^4}=-x^2$$

I don't understand how you arrived at this part.

******************************************
Do you agree that $(x^6+8)^{1/3} = (x^6)^{1/3} (1 + 8 x^{-6})^{1/3} = x^2 (1 + 8 x^{-6})^{1/3}$? If so, then do you agree that we can apply the general binomial expansion? This would be $(1+v)^r = 1 + rv + O(v^2)$, which holds for $|v| < 1$ and any real $r$ (positive, negative, rational, irrational---anything). If so, just let $v = 8/x^6$ and $r = 1/3$.
*******************************************

If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?
Ah, you're right. it should be MINUS root 3 because $$\sqrt{x^4}=-x^2$$

If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?

Can you link me the formula for that expansion? I've never seen it before.

Where did the O(v^2) come from? The formula seems to stop at 1+rv.

Ray Vickson
Homework Helper
Dearly Missed
Can you link me the formula for that expansion? I've never seen it before.
Google "binomial expansion". Alternatively, just apply the Maclauren (Taylor) series around $v=0$ to the function $f(v) = (1+v)^r$.

I won't be learning about power series until the end of next semester, so it's likely that your explanation is a bit above my level for the time being.

I'm still a bit confused by why it's - root 3.

The only example problem in my book defines (x^6)^1/2= - x^3, but when they factor $$\sqrt{25x^6}$$ it comes out to +5 rather than -5.

HallsofIvy
Homework Helper

## Homework Statement

Find $$lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}$$

N/A

## The Attempt at a Solution

Factoring out $$\frac {(-x^6)^{1/3}}{-x^2}$$ leaves me with $$\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2 }}$$
No, it doesn't. It leaves you with $$\frac{(-1-8x^{-6})}{-4- (3+x^{-4})^{1/2 }}$$

Taking the limit at infinity gives me $$\frac{-1}{-4+3^{1/2}}$$, which is wrong.

Ray Vickson
Homework Helper
Dearly Missed
Why doesn't the root 25 change sign in the following picture: http://s10.postimg.org/qb9mxcsmf/1243124.png

It's the same idea except that instead of $$\sqrt{x^4}=-x^2$$ we have $$\sqrt{x^6}=-x^3$$
We should NOT have
$$\sqrt{x^4} = -x^2 \:\longleftarrow\;\text{wrong!}$$
For any real $x$ we have $\sqrt{x^4} = x^2$ (because $\sqrt{x^4} > 0$ and $x^2 > 0$ if $x \neq 0$; this is the same, whether $x$ is positive or negative. However, $\sqrt{x^6}$ is different. Again, we always have $\sqrt{x^6} > 0$, but now $x^3$ is >0 if $x > 0$ and is < 0 if $x < 0$. In fact, if $x < 0$ then $x^3 < 0$, so we need $\sqrt{x^6} = -x^3$ in that case (where $x < 0$). On the other hand, if $x > 0$ then we have $\sqrt{x^6} = x^3$. To summarize:
$$\sqrt{x^4} = x^2 \;\; \text{always}\\ \sqrt{x^6} = \begin{cases} x^3 & \text{if} \;\; x > 0 \\ -x^3 & \text{if} \;\; x < 0 \end{cases}$$

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