Help to find this limit: limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

[kulki]

limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

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you can copy and paste to http://web2.0calc.com/ for see the answer and the question more appropriately

 

[Steven G]

limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

​

you can copy and paste to http://web2.0calc.com/ for see the answer and the question more appropriately

Hi, Sure we will help you find the limit. I guess you got distracted when you made the post since you failed to include where you need help/got stuck. Don't feel bad as it happens to all of us.

 

[kulki]

Hi, Sure we will help you find the limit. I guess you got distracted when you made the post since you failed to include where you need help/got stuck. Don't feel bad as it happens to all of us.

Can you help me ?

 

[Steven G]

Can you help me ?

Yes, and I will be even happy to help. What do you need help with? Please be precise so I can do the best possible job with helping you.

 

[kulki]

Yes, and I will be even happy to help. What do you need help with? Please be precise so I can do the best possible job with helping you.

Just finding this limit which goes to infinity

 

[lookagain]

... when you made the post since you failed to include where you need help/got stuck.

kulki, the post above is a prompt for you to type out and/or display images of steps up to where you understand.


And read this, too:

http://www.freemathhelp.com/forum/threads/41538-Read-Before-Posting!!

 

[Steven G]

Just finding this limit which goes to infinity

Let \(\displaystyle y =(\dfrac{x+2}{x-2})^{x+1}\)

Take ln of both sides. Then compute the limit of both sides as x goes to infinity.

Show us your work and we will go from there.

 

[kulki]

Let \(\displaystyle y =(\dfrac{x+2}{x-2})^{x+1}\)

Take ln of both sides. Then compute the limit of both sides as x goes to infinity.

Show us your work and we will go from there.

lny=(x+1).ln(x+2)/(x-2) then what is the next step?

 

[Steven G]

lny=(x+1).ln(x+2)/(x-2) then what is the next step?

Good. Now do what I said before, compute the limit as x approaches infinity of both sides (really just the right side).

 

[kulki]

Good. Now do what I said before, compute the limit as x approaches infinity of both sides (really just the right side).

ı did l'Hopital for ln(x+2)/(x-2) and found 0 ,now 0 times infinity what is next ?

 

[Steven G]

ı did l'Hopital for ln(x+2)/(x-2) and found 0 ,now 0 times infinity what is next ?

You only use l'Hopital if you have a product or a quotient and then under certain conditions. There is not a product or quotient in a single term like ln(x+2)/(x-2).

So tell us what exactly you did with ln(x+2)/(x-2) and l'Hopital rule.

 

[pka]

limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

\(\displaystyle {\large{\displaystyle\lim _{x \to \infty }}{\left( {1 + \frac{a}{{x + b}}} \right)^{cx + d}} = {e^{ac}}}\)

\(\displaystyle \begin{align*} \left(\dfrac{x+2}{x-2}\right)^{x+1}&=\left(1+\dfrac{4}{x-2}\right)^{x+1}\\&\therefore \to e^4 \end{align*}\)