Solving Limits that can't be rationalized (i think)

[hoaver]

Homework Statement

<br /> \lim_{x\to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)

<br /> \lim_{x\to 0}\left(\frac{2^2^x {-2^x}}{2^x{-1}}\right)

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?

 

[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

 

[hoaver]

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

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

Hopefully someone knows how to solve the 1st one

 

[gabbagabbahey]

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

 

[hoaver]

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

 

[gabbagabbahey]

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

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.

 

[icystrike]

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.

note I am not thread stater
when i apply l'Hôpital's i will get 0.75
however if i don't i get infinity

 

[gabbagabbahey]

note I am not thread stater
when i apply l'Hôpital's i will get 0.75
however if i don't i get infinity

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)?

 

[icystrike]

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)?

opps. my bad. miscalculated

 

[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.

 

[hoaver]

Thanks for the help everyone.

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

 

[Дьявол]

You did rationalization of the denominator.

Regards.

 

[Mentallic]

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

You multiplied by its conjugate.

 

[hoaver]

You multiplied by its conjugate.

Thats the one. Thanks =)