Question

Lim n-> infinity sqrt(n)/n^sin(n)

Answer 1

\[\lim_{n \rightarrow \infty} \sqrt{n}/n ^{\sin n}\]

Answer 2

Well in general, what does sin(x) do as x goes to infinity?

Answer 3

sin(x) is bounded between -1 and 1 but I am not sure exactly what value it approaches

Answer 4

@missinglink what is the answer?0?

Answer 5

The answer to the limit? I don't have the answer.

Answer 6

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.
\[\Huge \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n ^{\sin n}}=\frac{\frac{1}{2 \sqrt{n}}}{n^{\sin n}\log n}=\frac{0}{..}=0\]

Answer 7

Is the derivative of n^sin(n) = n^sin(n)*log(n)? How did you get that?

Answer 8

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.

Answer 9

\[\Huge \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n ^{\sin n}}=\frac{\frac{1}{2 \sqrt{n}}}{n^{\sin n}\log n \cos n}=\frac{0}{..}=0\]
i missed a term here
a^x=a^x loga

Answer 10

Basically what the graph does.