# Homework Help: Calculus: Limits help please

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).

21. Feb 25, 2005

### a_ng116

I stumbled upon another way to solve this question without using the derivative method since according to my teacher, the derivative method is supposely too advanced for our level. Anyways, putting in a 1 in the bottom so it becomes:
$$\frac {\sqrt[3]{1+x^2} - 1}{(1+x^2)-1}$$

and then taking a difference of cubes on the bottom it becomes:

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

The top and bottom divide to give 1 and putting a 0 for x $$\frac{1}{(\sqrt[3]{1+x^2})^2 + \sqrt[3]{1+x^2} + 1)}$$ it gets you $$\frac {1}{3}$$ also.Apparently,this was the way my teacher wanted us to do it. Well, enough of my rambling. Thank you all for your patience and taking the time to assist me. Cheers.

22. Feb 27, 2005

### Galileo

This is exactly the same as substituting $t=\sqrt[3]{1+x^2}$, it's just easier to spot that: $(t^3-1)=(t-1)(t^2+t+1)$
instead of:
$$(1+x^2)-1=(\sqrt[3]{1+x^2} -1)(\sqrt[3]{1+x^2})^2 + \sqrt[3]{1+x^2} + 1)$$

[/end nonconstructive post]

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