Limit problem

1. Feb 5, 2015

ciubba

1. The problem statement, all variables and given/known data
Find $$lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}$$

2. Relevant equations
N/A

3. 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.

2. Feb 5, 2015

Quantum Defect

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)]

3. Feb 5, 2015

Svein

Your arithmetic is wrong (in the denominator).

4. Feb 5, 2015

Ray Vickson

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 $(3 x^4 + 1)^{1/2}$?

5. Feb 5, 2015

ciubba

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.

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?

6. Feb 5, 2015

Quantum Defect

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?

7. Feb 5, 2015

8. Feb 5, 2015

ciubba

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

9. Feb 5, 2015

Quantum Defect

10. Feb 5, 2015

ciubba

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

11. Feb 5, 2015

Ray Vickson

Google "binomial expansion". Alternatively, just apply the Maclauren (Taylor) series around $v=0$ to the function $f(v) = (1+v)^r$.

12. Feb 5, 2015

ciubba

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.

13. Feb 5, 2015

HallsofIvy

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

14. Feb 5, 2015

ciubba

15. Feb 5, 2015

Ray Vickson

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:
$$\sqrt{x^4} = x^2 \;\; \text{always}\\ \sqrt{x^6} = \begin{cases} x^3 & \text{if} \;\; x > 0 \\ -x^3 & \text{if} \;\; x < 0 \end{cases}$$

Last edited: Feb 6, 2015