Find the limit of lim x-> 0 x^2 sin (1/x) cos (1/x)

[kimmy_koo51]

Find the limit of lim x^2 sin (1/x) cos (1/x)
x-> 0

I have no idea where to go with this one.

 

[o_O]

Looking at the sin(1/x)cos(1/x) reminds me of the identity:
sinA = 2sinAcosA \(\displaystyle \to\) (sinA)/2 = sinAcosA where A = 1/x here. So your expression could be expressed as:

\(\displaystyle \lim_{x \to 0} \left(\frac{1}{2} x^{2} sin (\frac{2}{x}) \right)\)

=\(\displaystyle \frac{1}{2} \cdot \lim_{x \to 0} \left( x^{2} sin (\frac{2}{x}) \right)\)

Now my question is if you recognize what the limit of this is:
\(\displaystyle \lim_{x \to 0} \left(x^{2}sin(\frac{1}{x}) \right)\)

Edit: For that last limit I gave you, it involves using the squeeze theorem. Once you figure out what the limit is, then with a bit of tweaking you could find the limit of YOUR question.

 

[pka]

Find the limit of lim x^2 sin (1/x) cos (1/x)
x-> 0

Just note that:
\(\displaystyle \L
\left| {x^2 \sin \left( {\frac{1}{x}} \right)\cos \left( {\frac{1}{x}} \right)} \right| = \left| {x^2 } \right|\left| {\sin \left( {\frac{1}{x}} \right)\cos \left( {\frac{1}{x}} \right)} \right| \le \left| {x^2 } \right|\)