Exploring the Limits: Evaluating a Complex Fraction

[ciubba]

Homework Statement

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

Homework Equations

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.

 

[Quantum Defect]

Homework Statement

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

Homework Equations

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 Statement

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

Homework Equations

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

 

[ciubba]

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]

[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]

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?

 

[ciubba]

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

 

[Quantum Defect]

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

http://en.wikipedia.org/wiki/Binomial_approximation

 

[ciubba]

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

 

[Ray Vickson]

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 $f(v) = (1+v)^r$.

 

[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.

 

[HallsofIvy]

Homework Statement

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

Homework Equations

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.

 

[ciubba]

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 $$\sqrt{x^6}=-x^3$$

 

[Ray Vickson]

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