1. Feb 22, 2005

### a_ng116

1) $$\lim_{x\rightarrow 0}\frac {\sqrt[3]{1+x^2} -1}{x^2}$$

I tried doing a difference of cubes to the top and I got: $$\frac{\sqrt[3]{1+x^2}-1 ((\sqrt[3]{1+x^2})^2 + \sqrt[3]{1+x^2} + 1)} {x^2}$$

I know you need to get rid of the $$x^2$$ on the bottom of the equation but now I'm stuck. Am I approaching this the right way and does anyone have any suggestions? Please and thank you.

Last edited: Feb 22, 2005
2. Feb 22, 2005

### xanthym

L'Hospital's Rule applies:

$$:(1): \ \ \ \ \lim_{x\rightarrow 0}\frac {\sqrt[3]{1+x^2} -1}{x^2}$$

$$:(2): \ \ \ \ = \ \lim_{x\rightarrow 0}\frac {(1/3)(2x)(1+x^2)^{-2/3}} {2x}$$

$$:(3): \ \ \ \ = \ \lim_{x\rightarrow 0} {(1/3)(1+x^2)^{-2/3}}$$

$$:(4): \ \ \ \ = \ (1/3)$$

~~

3. Feb 23, 2005

### himanshu121

you can use binomial expansion for

$$({1+x^2})^{\frac{1}{3}}$$

4. Feb 23, 2005

### Galileo

Make the subsitution $h = x^2$.

$$\lim_{h\rightarrow 0}\frac {\sqrt[3]{1+h} -1}{h}$$

Does the form of the limit remind you of something familiar?
Looks like the derivative of a function at some point...

Alternatively, you could use the sub: $t=\sqrt[3]{1+x^2}$

Last edited: Feb 23, 2005
5. Feb 23, 2005

### a_ng116

I thought I got it but I didn't. $$\lim_{h\rightarrow 0}\frac {\sqrt[3]{1+h} -1}{h}$$ looks like is from the formula $$\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$. Or perhaps not? Could anyone give me another hint or suggestion involving the question using the method above,replacing $$x^2$$ with $$h$$ please? Thank you very much.

6. Feb 23, 2005

### Justin Lazear

You're on the right track. It does look a lot like that. Now what's f(x) and what's a? Note that $f(a + h) = \sqrt[3]{1 + h}$ and $f(a) = 1$.

--J

7. Feb 24, 2005

### a_ng116

So working backwards I got $${\sqrt[3] {x} }$$ . Using the formula, replacing x with a $$1+h$$ then: $$\frac {\sqrt[3] {1+h} - \sqrt[3] 1} {h}$$ which is $$\lim_{h\rightarrow 0}\frac {\sqrt[3]{1+h} -1}{h}$$. Going back to $$\lim_{h\rightarrow 0}{\sqrt[3] {x} }$$ and replacing x with 0 gives you 0. I don't think I'm seeing it yet. Do you replace the $${\sqrt[3] {x} }$$ back some point into the equation?

8. Feb 24, 2005

### Justin Lazear

Nope. You know that the limit is the derivative of the function evaluated at the point a. f(x) indeed equals x1/3. I think you'll agree with me in saying that calculating the derivative of x1/3 is pretty easy, yes? All you have to do is evaluate the derivative of it at a.

Have you figured out what a is?

--J

9. Feb 24, 2005

### a_ng116

could a be 1?

10. Feb 24, 2005

### Curious3141

Yes, a is 1.

I like Galileo's method, it shows insight.

11. Feb 24, 2005

### a_ng116

so using Galieo's method, do i replace x with 1 so $$\sqrt[3]1=1$$ ?

12. Feb 24, 2005

### dextercioby

The method with the derivatives is interesting/ingenious,but the one using
$$A^{3}-B^{3}=(A-B)(A^{2}+AB+B^{3})$$

is simply elegant...

So the OP needs to redo calculations...

Daniel.

13. Feb 24, 2005

### Galileo

The substitution $t=\sqrt[3]{1+x^2}$ is easiest I think.
The limit becomes:

$$\lim_{t\to 1}\frac{t-1}{t^3-1}$$

De derivative method is cute and works sometimes, but usually a limit of this type shows up exactly when you want to calculate the derivative, so you cannot assume you know what the derivative of $\sqrt[3]{x}$ is.
The same is true for l'hospitals method.

14. Feb 24, 2005

### Justin Lazear

You need to differentiate f(x) before you set x to a. Remember that the equation is f'(x)|x=a equals the limit, not f(a).

--J

Last edited: Feb 24, 2005
15. Feb 24, 2005

### dextercioby

It's more like:
$$f'(a)=f'(x)|_{x=a}$$ and NOT $$f'(a)=[f(a)]' (=0)$$

Daniel.

16. Feb 24, 2005

### Justin Lazear

Yes, of course. Must spend more time composing posts.

--J

17. Feb 24, 2005

### a_ng116

So by using this method, taking a difference of cubes on the bottom, the limit as t approaches 1 is $$\frac {1}{3}$$ which is the same answer as by using L'Hospital's Rule such as that in post #2. I'm sorry in advance for my stupidity but the bottom (which was originally $$x^2$$ ) how did it become $$t^3-1$$ ?

18. Feb 24, 2005

### Nx2

i dunno... but can't u just rationalize this limit?

- Tu

19. Feb 24, 2005

### Justin Lazear

Solve for x2 in the equation $t = \sqrt[3]{1+x^2}$ and substitute it into the denominator.

--J

20. Feb 25, 2005

### Galileo

Since the answer is known anyway:

if $t=\sqrt[3]{x^2+1}$, then $x^2=t^3-1$.

For the derivative method.
After the substitution $h=x^2$ the limit becomes:
$$\lim_{h \to 0}\frac{\sqrt[3]{1+h}-1}{h}$$

Now the derivative of $\sqrt[3]{x}$ at the point x=1 is:
$$\lim_{h \to 0}\frac{\sqrt[3]{1+h}-\sqrt[3]{1}}{h}$$
which is exactly you limit.
So the limit is equal to the value of the derivative of $\sqrt[3]{x}$ at the point x=1. The derivative is: $\frac{1}{3\sqrt[3]{x}^{2}}$, so at x=1 it's 1/3.
(Seeing it as the derivative of $\sqrt[3]{x+1}$ in the point x=0 works too).