Question

How would I evaluate this limit?

Answer 1

\[\LARGE \lim_{x \rightarrow 0} x^9 cos(5/x)\]

Answer 2

It is zero

Answer 3

\[
\left | x^9 \sin(5/x) \right|\le |x^9|
\]

Answer 4

So show it by squeeze theorem?

Answer 5

yes

Answer 6

\[
\left | x^9 \cos(5/x) \right|\le |x^9|
\]

Answer 7

I put sin instead of cos in my first post

Answer 8

let f(x)=x^9
g(x)=cos(5/x)
\[\lim_{x \rightarrow 0}f(x)=\lim_{x \rightarrow 0}x ^{9}=0,\left| g(x) \right|=\left| \cos \left( \frac{ 5 }{ x } \right) \right|\le 1\]
by squeeze principle
\[\lim_{x \rightarrow 0}f \left( x \right)g \left( x \right)=0\]

Answer 9

\[cos(u)=1-\frac{u^2}{2!}+\frac{u^4}{4!}-\frac{u^6}{6!}+...\]
\[cos(a/x)=1-\frac{(a/x)^2}{2!}+\frac{(a/x)^4}{4!}-\frac{(a/x)^6}{6!}+...\]
\[x^9~cos(a/x)=x^9-\frac{ax^7}{2!}+\frac{ax^5}{4!}-\frac{ax^3}{6!}+...\]
hmm .... just wondering lol

Answer 10

cos (a/x) is defined in deleted neighbourhood of 0