Question

compute lim x->0 (x^2(sin(4/x)))

Answer 1

\[\lim_{x \rightarrow 0} x^2 \sin(4/x)\]

Answer 2

Show steps please

Answer 3

how about squeeze theorem

Answer 4

The sine function is confined to [-1, 1]. Therefore,\[
-\lim_{x \rightarrow 0} x^2 ~~\le ~~ \lim_{x \rightarrow 0} x^2 \sin(4/x) ~~\le ~~ \lim_{x \rightarrow 0} x^2 \\
-0 ~~\le ~~ \lim_{x \rightarrow 0} x^2 \sin(4/x) ~~\le ~~ +0\\
\lim_{x \rightarrow 0} x^2 \sin(4/x) = 0
\]

Answer 5

pretty

Answer 6

thanks.

Answer 7

is that how I show that the limit equals 0 on an exam?

Answer 8

it depends on how picky your instructor is...look at the following for an example

http://openstudy.com/users/zarkon#/updates/54825bcfe4b0edc680fcf779

and look at what loser66 got when he did a squeeze theorem problem

Answer 9

I can add two more steps to my earlier reply just for clarification.

The sine function is confined to [-1, 1]. Therefore,\[
x^2*(-1) ~~\le ~~ x^2 * \sin(4/x) ~~\le ~~ x^2 * (1) \\
-x^2 ~~\le ~~ x^2 * \sin(4/x) ~~\le ~~ x^2\\
\text{ } \\
\lim_{x \rightarrow 0} -x^2 ~~\le ~~ \lim_{x \rightarrow 0} x^2 \sin(4/x) ~~\le ~~ \lim_{x \rightarrow 0} x^2 \\
-0 ~~\le ~~ \lim_{x \rightarrow 0} x^2 \sin(4/x) ~~\le ~~ +0\\
\lim_{x \rightarrow 0} x^2 \sin(4/x) = 0
\]