lim(n->infinity) sin(1/n) n^2/(2n-1)

[dagr8est]

lim(n->infinity) sin(1/n) n^2/(2n-1)

I don't see how to manipulate this expression to get the answer 1/2. Anyway I try it, I get the indeterminate form infinity(0). Any help is appreciated.

 

[skeeter]

lim(n->infinity) sin(1/n) n^2/(2n-1)

is equivalent to ...

lim{1/n -> 0} sin(1/n)*n/[(1/n)(2n-1)] =

lim{1/n -> 0} [sin(1/n)/(1/n)]*[n/(2n-1)] =

lim{1/n -> 0} [sin(1/n)/(1/n)] * lim{1/n -> 0} n/(2n-1) =

lim{1/n -> 0} [sin(1/n)/(1/n)] * lim{n -> inf} n/(2n-1) =

1 * 1/2 = 1/2

 

[galactus]

Exapnd your function:

\(\displaystyle \L\\\frac{sin(\frac{1}{n})n^{2}}{2n-1}=\frac{sin(\frac{1}{n})}{4(2n-1)}+\frac{nsin(\frac{1}{n})}{2}+\frac{sin(\frac{1}{n})}{4}\)

Now, \(\displaystyle \L\\\frac{nsin(\frac{1}{n})}{2}\) is the only limit which does not tend to 0. Guess what it is?.

We get this from making the observation that \(\displaystyle \L\\\lim_{n\to\infty}nsin(\frac{1}{n})=1\)

Let \(\displaystyle t=\frac{1}{n}\), then \(\displaystyle n=\frac{1}{t}\) and \(\displaystyle x\rightarrow\;\ {+\infty}\) as \(\displaystyle t\rightarrow \;\ 0^{+}\)
So, \(\displaystyle \L\\\lim_{n\to\infty}nsin(\frac{1}{n})=\lim_{t\to\0^{+}}\frac{sint}{t}=1\)

 

[dagr8est]

Hmm... I see. Thanks.