Question

how do I figure out the limit of n approaching infinity e^3/n+8

Answer 1

\[\lim_{n \rightarrow \infty} \frac{ e ^{3} }{n+8 }\]
One way to do this is to plug in infinity into n. Infinity + 8 is still essentially infinity. e^3 is a constant. Any constant divided by infinity is 0. (Imagine dividing a number by a very large number: the answer will be very small, and almost zero.)
There are a few other ways but this is the simplest. Hope it helps!

Answer 2

no it is e raise to the 3/n+8 that is what is throwing me off

Answer 3

So the equation is:
\[\lim_{n \rightarrow \infty}e ^{\frac{ 3 }{ n+8 }}\]
?

Answer 4

yes

Answer 5

Well, we can still plug infinity into n, and using the same logic, we have:
\[\lim_{n \rightarrow \infty} e ^{\frac{ 3 }{ \infty +8 }}\]
\[\lim_{n \rightarrow \infty} e ^{\frac{ 3 }{ \infty }}\]
\[\lim_{n \rightarrow \infty} e ^{0}\]

Does this help?

Answer 6

It's the same as \(\lim_{h\rightarrow0}e^h\) with \(h = \frac3{n+8}\).

Answer 7

yes hold on...let me double check because I believe I tried that and webassign did not like the answer.

Answer 8

thanks wotseit. I really dislike webassign. ugh. But that is correct...thanks

Answer 9

No problem! Glad I could help :)

Answer 10

the last step is change
\[e^0 \]
to a number for the answer.

Answer 11

You could also work with the logarithm of the expression.

Answer 12

It wouldn't hurt to remember your Order of Operations. You could have written:

e^[3/(n+8)]

Use parentheses to clarify intent.