Question

Help with limits?

Answer 1

How would I find the limit?\[\lim_{x \rightarrow 0} x^2 \sin \left( \frac{ 1 }{ x^2 } \right)\]

Answer 2

DNE

Answer 3

really?

Answer 4

hint: let u = 1/(x^2)

solve for x^2 to get x^2 = 1/u

therefore,

\[
\large
\lim_{x \rightarrow 0} x^2 \sin \left( \frac{ 1 }{ x^2 } \right)
=
\lim_{u \rightarrow \infty} \frac{1}{u}\sin \left( u \right)
=
???
\]

as x gets closer to 0, u = 1/(x^2) gets closer to positive infinity

Answer 5

sine is bounded by -1 and 1
x is going to zero

Answer 6

you can use old mr "squeeze box theorem" or whatever it is called

Answer 7

\[-x^2\leq x^2\sin(\frac{1}{x})\leq x^2\]

Answer 8

sorry about the DNE

Answer 9

lol "this channel does not exist" i guess that is where the DNE comes from ...

Answer 10

Jim: are you using the chain rule? Would that limit be 0?

Answer 11

:D @misty1212

Answer 12

no, the chain rule is for derivatives

Answer 13

and yes,

\[
\large
\lim_{x \rightarrow 0} x^2 \sin \left( \frac{ 1 }{ x^2 } \right)
=
\lim_{u \rightarrow \infty} \frac{1}{u}\sin \left( u \right)
=
0
\]

Answer 14

sin(u) is bounded on the interval [-1,1]

the denominator grows without bounds

so you effectively have 1/x as x goes to infinity. But now that I'm looking at this, I realize that we could also have -1/x as x goes to infinity.

Despite this, either 1/x or -1/x will get so small that they will both approach 0