Calculus Help for College Newbie: Limit to Infinity

[kuskus94]

Hi, I am a newbie here and a newbie too at calculus since I just started my first year in college. Anyway, there are some questions that I don't quite understand and I hope that anyone can help me with this.

http://sphotos-a.ak.fbcdn.net/hphotos-ak-ash4/269471_368130629940171_945927165_n.jpg

I am sorry for the use of an image because I think that the questions are quite complex for me to write on keyboard.

Thank you in advance. :)

 

[lurflurf]

a) write
$$\frac{\sqrt{4-x^2}}{x-2}=\sqrt{-1+\frac{4}{x-2}}$$
b) use the fact
$$e=\lim_{x \rightarrow \infty} \left( 1+\frac{1}{n} \right) ^n$$

 

[SammyS]

Hi, I am a newbie here and a newbie too at calculus since I just started my first year in college. Anyway, there are some questions that I don't quite understand and I hope that anyone can help me with this.

http://sphotos-a.ak.fbcdn.net/hphotos-ak-ash4/269471_368130629940171_945927165_n.jpg

I am sorry for the use of an image because I think that the questions are quite complex for me to write on keyboard.

Thank you in advance. :)

Hello kuskus94. Welcome to PF !

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

 

[kuskus94]

a) write
$$\frac{\sqrt{4-x^2}}{x-2}=\sqrt{-1+\frac{4}{x-2}}$$
b) use the fact
$$e=\lim_{x \rightarrow \infty} \left( 1+\frac{1}{n} \right) ^n$$

Dear lurflurf,

I sill don't get how the question a) can me modified to be like that and I would like to ask you to break it down more for me.

As for question b), I apologize that my lack of reading. It is my bad to not read the calculus book deeper.

Thank you for your reply.

Hello kuskus94. Welcome to PF !

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

Dear SammyS,

I thank you for welcoming me in this forum. But then I got a waring message from Mark44 (the admin to my suppose) beacuse I didn't state that I have tried doing this question. Well, it was my bad .

For the question a), I still got $${\sqrt{0}}$$ over 0. I don't think I have any other idea to workaround this question. I would be really glad if you can help me more.

Thank you in advance.

 

[SammyS]

...

Dear SammyS,

I thank you for welcoming me in this forum. But then I got a waring message from Mark44 (the admin to my suppose) because I didn't state that I have tried doing this question. Well, it was my bad .

For the question a), I still got $${\sqrt{0}}$$ over 0. I don't think I have any other idea to workaround this question. I would be really glad if you can help me more.

Thank you in advance.

You can cancel one of the factors in the numerator. What is $\displaystyle \frac{\sqrt{u}}{u}\$ or $\displaystyle \frac{\sqrt{3}}{3}\ ?$

 

[kuskus94]

You can cancel one of the factors in the numerator. What is $\displaystyle \frac{\sqrt{u}}{u}\$ or $\displaystyle \frac{\sqrt{3}}{3}\ ?$

Ah, I see... Thank you very much for your help. :)