anonymous
  • anonymous
lim as x approaches infinity (1+3/x)^(5x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
Okay well it says that the answer is e^15 .. how is that?!
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty} (1+{3\over x}^{5x})\]is that what it is?

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anonymous
  • anonymous
ops, typo
anonymous
  • anonymous
the exponent should be out of the parenthesis
anonymous
  • anonymous
Yes that is right besides the exponent outside of the parenthesis
anonymous
  • anonymous
hmmm....definitely looks like 1 to me
anonymous
  • anonymous
Yeah I wasn't sure how they got e^15. I don't understand at all
anonymous
  • anonymous
the answer is e^15 but i'm not sure how to get there
myininaya
  • myininaya
e^(5/3)
myininaya
  • myininaya
i mean e^(15) sorry
anonymous
  • anonymous
how did yu get there myininaya ?
myininaya
  • myininaya
recall limit x->infinity (1+1/x)^x=e
anonymous
  • anonymous
Is that something I should know off the top of my head?
myininaya
  • 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
  • anonymous
oh boy.. i never liked limits anyways.. lol
anonymous
  • anonymous
good catch myininaya thats just one of those weird math things
anonymous
  • anonymous
Okay I remember the sinx/x = 1
anonymous
  • anonymous
So where did the 15 come from
anonymous
  • anonymous
from the 3/x instead of 1/x and the power of 5x instead of x
myininaya
  • myininaya
http://answers.yahoo.com/question/index?qid=20110223065841AAHpCSc
myininaya
  • 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
  • anonymous
Okay that makes sense now
anonymous
  • anonymous
Thank you so much!
myininaya
  • myininaya
np
myininaya
  • 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
  • anonymous
Which is why it comes to e in the above limits?
myininaya
  • 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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