missinglink
  • missinglink
Lim n-> infinity sqrt(n)/n^sin(n)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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missinglink
  • missinglink
\[\lim_{n \rightarrow \infty} \sqrt{n}/n ^{\sin n}\]
Psymon
  • Psymon
Well in general, what does sin(x) do as x goes to infinity?
missinglink
  • missinglink
sin(x) is bounded between -1 and 1 but I am not sure exactly what value it approaches

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DLS
  • DLS
@missinglink what is the answer?0?
missinglink
  • missinglink
The answer to the limit? I don't have the answer.
DLS
  • DLS
Well.. Just an attempt if you don't have the answer. \[\Huge \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n ^{\sin n}}=\frac{\infty}{\infty}\] Therefore L-Hospital is valid. \[\Huge \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n ^{\sin n}}=\frac{\frac{1}{2 \sqrt{n}}}{n^{\sin n}\log n}=\frac{0}{..}=0\]
missinglink
  • missinglink
Is the derivative of n^sin(n) = n^sin(n)*log(n)? How did you get that?
Psymon
  • Psymon
Itdoesnt approach a value. It continues to oscillate up and down, never approaching a value. Nothing changes when we have sin(x) as a power or as a multplication. If you are unsure about what happens when sin(x) is an exponent, we can just use ln to make it a regular multiplication: \[y = \frac{ \sqrt{n} }{ n ^{\sin(n)} }\] \[\ln(y) = \ln(\frac{ \sqrt{n} }{ n ^{\sin(n)} }) \] \[\ln(y) = \frac{ 1 }{ 2 }\ln(n) - \sin(n)*\ln(n)\] Now we know that sin(n) oscillates. And right now, ln(n) is just a numerical multiplier of sin(n). A multiplier that would become infinity at that. So this multiplier would only cause sin(n) to osciallte more wildly than before. (So yeah, basically Im going to disagree with DLS here) In the end, we have infinity - and infinite multiplier of an osciallting function. So the limit does not exist, it oscillates as well.
DLS
  • DLS
\[\Huge \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n ^{\sin n}}=\frac{\frac{1}{2 \sqrt{n}}}{n^{\sin n}\log n \cos n}=\frac{0}{..}=0\] i missed a term here a^x=a^x loga
Psymon
  • Psymon
|dw:1378318497769:dw| Basically what the graph does.
Psymon
  • Psymon
That crook on the end isa mistake, lol.
missinglink
  • missinglink
@Psymon I kind of understand your explanation but we do end up getting some answer when we apply lhopitals rule right? I am leaning towards your answer but when I applied lhopitals rule like DNS mentioned, the limit seems to approach zero.
DLS
  • DLS
DLSSSSSSSSSSSSSSSSSSSSSSSSSSS
missinglink
  • missinglink
@DLS, Sorry about that.
DLS
  • DLS
It won't be 0 logically,@Psymon is right,it does not exist.
DLS
  • DLS
if you think logically its infinity/infinity(-1 to 1) that means if it is -1 then it becomes non 0 and if it is around 1 then it tends to 0,nothing can be said IMO.
missinglink
  • missinglink
Yes I think Psymon is right too but I guess I will ask in my class why lhopitals rule gives zero.

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