# limits

• October 22nd 2008, 03:12 PM
bigb
limits
Use the definition of a limit epsilon and delta to find the following limit.

• October 22nd 2008, 04:19 PM
Plato
Note that in the limit $x \ne 0$ thus $\left| {\sin \left( {\frac{1}{{x^2 }}} \right)} \right| \leqslant 1$
So $\left| {x\sin \left( {\frac{1}{{x^2 }}} \right)} \right| \leqslant \left| x \right|$.
Just let $\varepsilon > 0 \Rightarrow \quad \delta = \varepsilon$.
Now you use the definition.