lim as x approaches infinity (1+3/x)^(5x)

- anonymous

lim as x approaches infinity (1+3/x)^(5x)

- chestercat

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- anonymous

1 becuase as x approach infinity the 3/x goes to zero which just leaves 1^(infinity) and 1 to any power is 1

- anonymous

Okay well it says that the answer is e^15 .. how is that?!

- anonymous

\[\lim_{x \rightarrow \infty} (1+{3\over x}^{5x})\]is that what it is?

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## More answers

- anonymous

ops, typo

- anonymous

the exponent should be out of the parenthesis

- anonymous

Yes that is right besides the exponent outside of the parenthesis

- anonymous

hmmm....definitely looks like 1 to me

- anonymous

Yeah I wasn't sure how they got e^15. I don't understand at all

- anonymous

the answer is e^15 but i'm not sure how to get there

- myininaya

e^(5/3)

- myininaya

i mean e^(15) sorry

- anonymous

how did yu get there myininaya ?

- myininaya

recall limit x->infinity (1+1/x)^x=e

- anonymous

Is that something I should know off the top of my head?

- myininaya

i guess lol
i just recalled it though like a memory
or you know lim x->0 sinx/x=1
or some property like that

- anonymous

oh boy.. i never liked limits anyways.. lol

- anonymous

good catch myininaya
thats just one of those weird math things

- anonymous

Okay I remember the sinx/x = 1

- anonymous

So where did the 15 come from

- anonymous

from the 3/x instead of 1/x and the power of 5x instead of x

- myininaya

http://answers.yahoo.com/question/index?qid=20110223065841AAHpCSc

- myininaya

they show that lim x-> infinity (1+a/x)^x=e^a
so if we have lim x-> infinity (1+3/x)^x=e^3
and if we have lim x-> infinity (1+3/x)^(5x)=lim x->infinity [(1+3/x)^x]^5=
(the inside goes to e^3) = e^[3(5)]=e^15

- anonymous

Okay that makes sense now

- anonymous

Thank you so much!

- myininaya

np

- myininaya

catherine you can also see if we plug in really big values for x (1+1/x)^x is getting closer to e

- anonymous

Which is why it comes to e in the above limits?

- myininaya

yes you do that for any function
you can plug values in for whatever x is getting closer to to see where the function is getting closer to as x gets closer to whatever number its approaching

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