1. ## Help plz!

It's again prbls w/ l'Hospital rule.
omg, my calc prof. assigned like 50 prbls but i had to study for my other finals. these are due today at 9. plz help.

1) lim(1+e^x)^(1/x)
x->infinity
"infinity^0" type?
=e^lim (1/x)ln(1+e^x)

2)lim (1+(3/x^2)^x
x->infinity
"1^infinity" type
=e^ilm x ln(1+3/x^2)

3) lim x sin x/ (1-cos x)
this one i have no idea what to do.

plz help me so that i can just get some sleep. it's already 4 in the morning.

2. Originally Posted by katieeej
It's again prbls w/ l'Hospital rule.
omg, my calc prof. assigned like 50 prbls but i had to study for my other finals. these are due today at 9. plz help.

1) lim(1+e^x)^(1/x)
x->infinity
"infinity^0" type?
=e^lim (1/x)ln(1+e^x)

2)lim (1+(3/x^2)^x
x->infinity
"1^infinity" type
=e^ilm x ln(1+3/x^2)

3) lim x sin x/ (1-cos x)
this one i have no idea what to do.

plz help me so that i can just get some sleep. it's already 4 in the morning.
3) has already been done here: http://www.mathhelpforum.com/math-he...le-urgent.html Please pay closer attention so that you don't post the same questions more than once.

3. Originally Posted by katieeej
It's again prbls w/ l'Hospital rule.
omg, my calc prof. assigned like 50 prbls but i had to study for my other finals. these are due today at 9. plz help.

1) lim(1+e^x)^(1/x)
x->infinity
"infinity^0" type?
=e^lim (1/x)ln(1+e^x)

2)lim (1+(3/x^2)^x
x->infinity
"1^infinity" type
=e^ilm x ln(1+3/x^2)

3) lim x sin x/ (1-cos x)
this one i have no idea what to do.

plz help me so that i can just get some sleep. it's already 4 in the morning.
I will get you started with 1). 2) is similar and easier.

$\displaystyle \lim_{x \rightarrow +\infty} \left( 1 + e^x \right)^{\frac{1}{x}} = \lim_{x \rightarrow +\infty} \exp \left( \frac{\ln (1 + e^x)}{x}\right) = \exp \left( \lim_{x \rightarrow +\infty} \frac{\ln (1 + e^x)}{x} \right)$.

(The justification is left for you to give).

So you need to consider the limit $\displaystyle \lim_{x \rightarrow +\infty} \frac{\ln (1 + e^x)}{x}$. Use l'Hopital's Rule twice.

4. Originally Posted by mr fantastic
I will get you started with 1). 2) is similar and easier.

$\displaystyle \lim_{x \rightarrow +\infty} \left( 1 + e^x \right)^{\frac{1}{x}} = \lim_{x \rightarrow +\infty} \exp \left( \frac{\ln (1 + e^x)}{x}\right) = \exp \left( \lim_{x \rightarrow +\infty} \frac{\ln (1 + e^x)}{x} \right)$.

(The justification is left for you to give).

So you need to consider the limit $\displaystyle \lim_{x \rightarrow +\infty} \frac{\ln (1 + e^x)}{x}$. Use l'Hopital's Rule twice.
i'm so sorry but can you give me some more details?

5. Originally Posted by katieeej
i'm so sorry but can you give me some more details?
What don't you follow?

Edit: Note: $\displaystyle (1 + e^x)^{\frac{1}{x}} = e^{\ln (1 + e^x)^{\frac{1}{x}}} = e^{\frac{1}{x}\ln (1 + e^x)}$