# Homework Help: Solving Limits that can't be rationalized (i think)

1. Dec 10, 2008

### hoaver

1. The problem statement, all variables and given/known data
$$\lim_{x\to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)$$

$$\lim_{x\to 0}\left(\frac{2^2^x {-2^x}}{2^x{-1}}\right)$$

3. The attempt at a solution

I tried rationalizing but doesn't really help in evaluating these limits.

If a limit cannot be factored, what other ways are there of evaluating limits?

2. Dec 10, 2008

### Mentallic

For the first I can't seem to figure it out either. The denominator can be rationalised of course, but what comes after that I'm unsure.

As for the second, notice that $$2^{2x}=(2^x)^2$$ since $$(a^b)^n=a^{bn}$$
From there you can factorise and simplify

3. Dec 10, 2008

### hoaver

Ah, okay, I see how to solve the 2nd one now, thanks =).

Hopefully someone knows how to solve the 1st one

4. Dec 10, 2008

### gabbagabbahey

Have you learned L'Hopital's rule yet? If so, the first limit is fairly simple with that method.

5. Dec 10, 2008

### hoaver

We have not learned L'Hopital's rule or derivatives yet =(

6. Dec 10, 2008

### gabbagabbahey

In that case try multiplying both the numerator and denominator by $(2+\sqrt{x})(3+\sqrt{2x+1})$ and then simplify.....this should remove that pesky 0/0.

7. Dec 12, 2008

### icystrike

when i apply l'Hôpital's i will get 0.75
however if i dont i get infinity

8. Dec 12, 2008

### gabbagabbahey

0.75=3/4 is corrrect. You shouldn't be getting infinity using my method....what is $$\left(\frac{(2-\sqrt{x})(2+\sqrt{x})}{(3-\sqrt{2x+1})(3+\sqrt{2x+1})}\right) \left( \frac{3+\sqrt{2x+1}}{2+\sqrt{x}} \right)$$?

9. Dec 12, 2008

### icystrike

10. Dec 12, 2008

### Дьявол

$$\lim_{x \to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)*\frac{2+\sqrt{x}}{3+\sqrt{2x+1}}*\frac{3+\sqrt{2x+1}}{2+\sqrt{x}}$$

Or:

$$\lim_{x \to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)*\frac{2+\sqrt{x}}{3+\sqrt{2x+1}}*\frac{3}{2}$$

Regards.

11. Dec 12, 2008

### hoaver

Thanks for the help everyone.

Later, I also found it to work with using this:
$$\lim_{x \to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)*\frac{3+\sqrt{2x+1}}{3+\sqrt{2x+1}}$$
Simplifies to
$$\lim_{x \to 4}\left(\frac{(2-\sqrt{x})(3+\sqrt{2x+1})}{8-2x}}\right)$$
to
$$\lim_{x \to 4}\left(\frac{(2-\sqrt{x})(3+\sqrt{2x+1})}{2(2-\sqrt{x})(2+\sqrt{x})}}\right)$$

dividing out the 2-sqrt(x) results in:
$$\lim_{x \to 4}\left(\frac{3+\sqrt{2x+1})}{2(2+\sqrt{x})}\right)$$

after which I just subbed in 4 for the answer 3/4

edit: does anyone know what this method is called? multiplying by the ______ (what i did to the original equation)

Last edited: Dec 12, 2008
12. Dec 13, 2008

### Дьявол

You did rationalization of the denominator.

Regards.

13. Dec 13, 2008

### Mentallic

You multiplied by its conjugate.

14. Dec 13, 2008

### hoaver

Thats the one. Thanks =)