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**URGENT: Limits**

## Homework Statement

__PART A__Let f(x) be a function defined on an open interval containing c, and let L be a real number. Given:

[tex]\mathop {\lim }\limits_{x \to a} f\left( x \right) = L[/tex]

For each real [tex]\varepsilon > 0[/tex] , there exists a real [tex]\delta > 0[/tex] such that for all x:

[tex]0 < \left| {x - c} \right| < \delta

0 < \left| {f\left( x \right) - L} \right| < \varepsilon [/tex]

__PART B__Hence, evaluate exactly [tex]\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta }[/tex]

**2. The attempt at a solution**

I do not have the slightest idea on how to solve part A, however for part B my working is as follows (I am using L'hopital's Rule):

[tex]

\begin{array}{l}

\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta } = \frac{{\sin ^2 4\left( 0 \right)}}{0} = \frac{0}{0} \\

\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta } = \mathop {\lim }\limits_{\theta \to 0} \frac{{f\left( \theta \right)}}{{g\left( \theta \right)}}\quad \Rightarrow \quad f\left( \theta \right) = \sin ^2 4\theta \quad g'\left( \theta \right) = 1 \\

f\left( \theta \right) = \sin ^2 4\theta \\

\sin ^2 \theta = \frac{{1 - \cos 2\theta }}{2} \\

f\left( \theta \right)\sin ^2 4\theta = \frac{{1 - \cos 8\theta }}{2} \\

\end{array}

[/tex]

Am i doing this part correctly? Is there an easier way? Many thanks to all help and suggestions offered.

unique_pavadrin

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