Question

(Tricky One) Limit of (x/(sin(1/x))-x^2) as x goes to infinity ... Thanks !!!! =)

Answer 1

are u allowed t use LHR ?

Answer 2

i don't know what you mean by LHR Oo

Answer 3

o.O lol ... sorry u dont need it at all...

Answer 4

is that\[\frac{x}{\sin \frac{1}{x}}-x^2\]?

Answer 5

yes !! =)

Answer 6

how about series...can u use series?

Answer 7

hmm.. maybe taylor series .. we didn't learn any more series yet .. but i will be happy if you show me what you think =]

Answer 8

well let \(t=\frac{1}{x}\) so \(t\rightarrow 0\) and\[\frac{x}{\sin \frac{1}{x}}-x^2=\frac{1}{t\sin t}-\frac{1}{t^2}=\frac{t-\sin t}{t^2 \sin t}\]for \(t \approx0\) we have \(\sin t\approx t-\frac{t^3}{6}\) so\[\lim_{t \rightarrow 0} \frac{t-\sin t}{t^2 \sin t}=\lim_{t \rightarrow 0} \frac{t-(t-\frac{t^3}{6})}{t^2 (t-\frac{t^3}{6})}=\lim_{t \rightarrow 0} \frac{\frac{1}{6}}{1-\frac{t^2}{6}}=\frac{1}{6}\]

Answer 9

Great !!!!!! you solved my mystery !!!!! =]

Answer 10

:)