## anonymous 5 years ago lim as x approaches infinity (1+3/x)^(5x)

1. 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

2. anonymous

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

3. anonymous

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

4. anonymous

ops, typo

5. anonymous

the exponent should be out of the parenthesis

6. anonymous

Yes that is right besides the exponent outside of the parenthesis

7. anonymous

hmmm....definitely looks like 1 to me

8. anonymous

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

9. anonymous

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

10. myininaya

e^(5/3)

11. myininaya

i mean e^(15) sorry

12. anonymous

how did yu get there myininaya ?

13. myininaya

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

14. anonymous

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

15. 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

16. anonymous

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

17. anonymous

good catch myininaya thats just one of those weird math things

18. anonymous

Okay I remember the sinx/x = 1

19. anonymous

So where did the 15 come from

20. anonymous

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

21. myininaya
22. 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

23. anonymous

Okay that makes sense now

24. anonymous

Thank you so much!

25. myininaya

np

26. myininaya

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

27. anonymous

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

28. 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