# Math Help - limits

1. ## limits

Hi, okay so I have the following limit:

lim x-->infinity [1+(4/(x-2))]^x

so I don;'t know how to get rid of the power of x...? I am pretty sure that I have to put it into a l'hopitalable form but because of the power of x I am stuck!

Thank you
note: I triedwritting all this nicely with html codes but it said there aas an error sorry about that!

2. Hi mate

$1 = (x-2)/(x-2)$

=> $1+(4/(x-2)) = (x-2)/(x-2) + 4/(x-2) = (x+2)/(x-2)$

$(x+2)/(x-2) = e^{ln [(x+2)/(x-2)]}$

$[1+(4/(x-2))]^x = e^{x*ln [(x+2)/(x-2)]}$

Now you should be able to get the solution $e^4$, for x -> infty

Yours,
Rapha

3. okay sorry I am still a little confused.. why did you equate 1 to (X-2)/ (x-2) ... I mean are we not supposed to l'hopital it at some point?

Your answer is right so what you did is 100% correct I just don't understand it!

4. Originally Posted by thehabsonice
okay sorry I am still a little confused..
why did you equate 1 to (X-2)/ (x-2) ...

I wanted to get one term for 1+(4/(x-2)), that is why i equated to (x-2)/(x-2)

Originally Posted by thehabsonice
I mean are we not supposed to l'hopital it at some point?
yes, we are, unless there is another solution, but I don't know an alternative.
So

my last step was

I'm gonna show that x*ln [(x+2)/(x-2)] = 4, x to infty

$x*ln [(x+2)/(x-2)] = \frac{1}{(x)^{-1}}*ln((x+2)/(x-2))$

Derivatives:

$(\frac{1}{(x)^{-1}})' = -1/x^2$

[ ln((x+2)/(x-2)) ] ' = 4/((x + 2)(2 - x))

Using L'Hospital leeds to the problem

$- \frac{4x^2}{(x + 2)(2 - x)}$

Are you able to show that $- \frac{4x^2}{(x + 2)(2 - x)} = 4$, x to infty

?

Originally Posted by thehabsonice
Your answer is right so what you did is 100% correct I just don't understand it!
Thank you, this statement is very polite

Rapha

5. Hello,

Otherwise, substitute $t=x-2 \Rightarrow x=t+2$

So the limit becomes :

$\lim_{t\to\infty} \left(1+\frac 4t\right)^{t+2}=\lim_{t\to\infty} \left(1+\frac 4t\right)^t \cdot \left(1+\frac 4t\right)^2$

The first one is known : $\lim_{x\to\infty}\left(1+\frac ax\right)^x=e^a$
and the second one just goes to 1.