# Anyone see how to solve this without L'Hospitals?

1. Nov 19, 2006

### dontdisturbmycircles

$$\lim_{x->1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$$

I can do it with L'Hospitals, but my teacher said there is also a way to do it without. I tried all sorts of techniques such as multiplying by the conjugate of both the denom/numerator as well as taking $$\sqrt{x}$$ out of both the denom and numerator. I can't see what method my teacher sees that I don't..

I get 2/3 with L'Hospitals by the way.

2. Nov 19, 2006

### robphy

factor
(a change of variables may help)

3. Nov 19, 2006

### dontdisturbmycircles

Ahhh okay I see, thanks.

4. Nov 20, 2006

### dontdisturbmycircles

Can someone demonstrate how to solve this with a change of variables? I am usually fairly comfortable using a change of variables but I don't see how to apply it here.

For example I can solve lim_{x->inf}e^(-2x) or w/e by making t=-2x. But I don't see what to do in with the problem in the thread.

I can solve it by dividing, but don't see how to use a change of variables.

Last edited: Nov 20, 2006
5. Nov 20, 2006

### robphy

If you can finish the problem by simply dividing, then you don't need the change of variables. My suggestion to change the variables may help you see how to write the numerator and the denominator so that each can be factored and have some common factors cancel, leaving you with something that you can easily take the limit of.

So, can you see how the numerator and denominator can be separately factored? (Think of polynomials.)

6. Nov 20, 2006

### dontdisturbmycircles

I don't see how to factor them :(. I can rewrite it as t=x-1 so it becomes

$$\lim_{t->0}\frac{t^{1/3}}{t^{1/2}}$$ but that screws it up.

Of course I could simplify to $$t^{-1/6}$$ but I screwed something up somewhere.

Last edited: Nov 20, 2006
7. Nov 20, 2006

### robphy

Let $$y=\sqrt[6]{x}$$.

8. Nov 20, 2006

### dontdisturbmycircles

$$\lim_{t->1}\frac{t^{2}-1}{t^{3}-1}$$
$$=\lim_{t->1}\frac{(t-1)(t+1)}{(t-1)(t^2+t+1)}$$
$$=\lim_{t->1}\frac{t+1}{t^{2}+t+1}$$
$$=\frac{2}{3}$$

Thanks alot :) I am still trying to understand your logic for using $$\sqrt[6]{x}$$ though.

9. Nov 20, 2006

### dontdisturbmycircles

I guess it makes sense that given x^n/x^m, replacing one with t^(nm) will make it easier to factor. I still have to think about it a bit.

10. Nov 20, 2006

$$\sqrt[6]{x} = t$$

$$t^{2} = \sqrt[6]{x}^{2} = x^{\frac{1}{3}}$$

$$t^{3} = \sqrt{x}$$

11. Nov 20, 2006

### robphy

I wanted to get the first line in your previous post... because I knew that I could factor out a common (t-1) factor from a term like (t^k-1). So, I wanted to choose of convenient change of variables get the numerator and denominator to be in the form (t^k-1).

Alongside the above, I wanted to eliminate all of the fractional powers of x.

12. Nov 20, 2006

### dontdisturbmycircles

Ah okay that makes perfect sense. Thanks alot!

13. Nov 20, 2006

### NateTG

L'Hospital's method can be derived from epsilon-delta definitions. Consider:

Assuming
$$x \neq 0$$
We have:
$$\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim_{x \rightarrow 0} \frac{\sqrt[3]{0+x}-1}{\sqrt{0+x}-1}=\frac{\frac{\sqrt[3]{0+x}-1}{x}}{\frac{\sqrt{0+x}-1}{x}}$$