Question

lim((n+5)/(n+3))^n n-> infinity ..how is the answer e^2

Answer 1

write the top as n+3+2

Answer 2

ok

Answer 3

then you have (1+2/(n+3))^n

Answer 4

let me give you the definition of e as a limit:
\[\lim_{n \rightarrow \infty}(1+\frac{r}{n})^n=e^r\]

Answer 5

ok

Answer 6

so now you have it in that form and you can see r=2

Answer 7

so it is e^2

Answer 8

ohhh do we use this def everytime we get a ques like this?

Answer 9

if you can put it into that form you can use the definition

Answer 10

Anytime you can make it fit that form. If not, you can use l'hospital's rule because you have infinity/infinity

Answer 11

raised to the infinity xDDDD

Answer 12

can you please see the attachment and explain the solution

Answer 13

thats how we get the definition of e as a limit

Answer 14

The way wolfram does it is using L'hospital's rule. When you do this, you realize that you have (inf/inf)^(inf) but you need to rewrite it so it fits a form that you can do l'hospital's on it. So you take the ln of it. So you get \[\ln(\lim_{n \rightarrow \infty} \frac{(n+5)}{(n+3)}^n)\]. Then you can bring down the n since its a power in a ln.
Then you write it as:
\[\lim_{n \rightarrow \infty} \frac{\ln(\frac{(n+5)}{n+3})}{\frac{1}{n}}\].
Then do l'hospital's. And evaluate. But remember, since you took a ln. You have to exponentiate it after you evaluate it. So, for this problem, the limit should evaluate to 2. Then you exponentiate it giving e^2

Answer 15

\[\sum_{1}^{\infty} 1/(2n+1) (2n-1)\]

Answer 16

Okay, that is a telescopic series. You need to do a partial fraction decomp. From there, expand out your terms until you get cancellation (find the nth partial sum) then take the lim. Let me know if you need it shown out.