Prove the following limit of a function using the definition of a limit (where c>0)
$\displaystyle \lim_{x\rightarrow{c}}{\frac{1}{x^3}} = \frac{1}{c^3} $
In response to the reply,no formula for delta depending on epsilon or c was given.For completeness sake,i will state the definition.
DEFINITION
Let $\displaystyle A \subseteq \mathbb{R}$, and let c be a cluster point of $\displaystyle A$. For a function $\displaystyle f : A \rightarrow{\mathbb{R}}$, a real number $\displaystyle L$ is said to be a limit of $\displaystyle f$ at c if, given any $\displaystyle \epsilon > 0$ there exists a $\displaystyle \delta > 0$ such that if $\displaystyle x \in A$ and $\displaystyle 0 < \mid x - c \mid < \delta$ then $\displaystyle \mid f(x) - L \mid < \epsilon$.
QUESTION
Prove the following limit of a function using the (above) definition where
c > 0
$\displaystyle \lim_{x \rightarrow{c}}{\frac{1}{x^3}} = \frac{1}{c^3}$