Thread: limit

1. limit

Show your work and calculate limit of the following questions.

2. For the first problem, which do you mean?
$\displaystyle \lim_{x \to \infty} 3 \sqrt{n}^\frac {1} {2n}$
or
$\displaystyle \lim_{x \to \infty} 3 \sqrt{n}^{\frac {1} {2} n}$

2nd problem

$\displaystyle \lim_{x \to \infty} (n+1)^\frac {1} {ln(n+1)}$
Take the ln of the lim, just remember to rise it to e later.
$\displaystyle \lim_{x \to \infty} \frac {ln(n+1)} {ln(n+1)}$

3. the first way you wrote it. and how can i just take the ln and do what you said for the second problem

4. I took the ln for you in problem 2...I guess i'll show it to you step by step

$\displaystyle \lim_{x \to \infty} (n+1)^\frac {1} {ln(n+1)}$
Just remember to rise the answer to e.
$\displaystyle \lim_{x \to \infty} ln(n+1)^\frac {1} {ln(n+1)}$
By the law of lns..
$\displaystyle \lim_{x \to \infty} \frac {1} {ln(n+1)}ln(n+1)$
$\displaystyle \lim_{x \to \infty} \frac {ln(n+1)} {ln(n+1)}=1$
So the answer will be $\displaystyle e^1$ or just plain e.

Try the first problem this way. You will probably need to l'hospital it.

5. Originally Posted by Linnus
I took the ln for you in problem 2...I guess i'll show it to you step by step

$\displaystyle \lim_{x \to \infty} (n+1)^\frac {1} {ln(n+1)}$
Just remember to rise the answer to e.
$\displaystyle \lim_{x \to \infty} ln(n+1)^\frac {1} {ln(n+1)}$
By the law of lns..
$\displaystyle \lim_{x \to \infty} \frac {1} {ln(n+1)}ln(n+1)$
$\displaystyle \lim_{x \to \infty} \frac {ln(n+1)} {ln(n+1)}=1$
So the answer will be $\displaystyle e^1$ or just plain e.

Try the first problem this way. You will probably need to l'hospital it.
cant use lhospital rule, so i dont know how to do it. Thats the problem i am having

6. Have you tried it? I just did and you don't actually need l'hospital.

$\displaystyle \lim_{n \to \infty} 3 \sqrt{n}^\frac {1} {2n}$

$\displaystyle 3 \lim_{n \to \infty} \sqrt{n}^\frac {1} {2n}$

$\displaystyle 3 \lim_{n \to \infty} \frac {ln\sqrt {n}} {2n}$

It should be obvious from here. Remeber just rise the limit to e, not including the 3.

$\displaystyle 3 \lim_{n \to \infty} \frac {\ln {n}} {4n}$

7. Originally Posted by Linnus
Have you tried it? I just did and you don't actually need l'hospital.

$\displaystyle \lim_{x \to \infty} 3 \sqrt{n}^\frac {1} {2n}$

$\displaystyle 3 \lim_{x \to \infty} \sqrt{n}^\frac {1} {2n}$

$\displaystyle 3 \lim_{x \to \infty} \frac {ln\sqrt {n}} {2n}$

It should be obvious from here. Remeber just rise the limit to e, not including the 3.

$\displaystyle 3 \lim_{x \to \infty} \frac {\ln {n}} {4n}$

I figuered it out. Its really quite simple. Just take the take the limit of the product so we have limit of 3^1/2n *limit of ((n)^(1/2))^(1/2n)

the limit of the first part is 1 since 1/2n will approch o and 3^0 is 1 and then same thing for the second part so 1*1 is 1