Math Help - limit of a function!

1. limit of a function!

find a \delta such that for the given value of L and \varepsilon in the statement \midf(x)-L\mid\le\varepsilon whenever 0\le\midx-a\mid\le\delta
given that f(x)=(x^2-9)/(x+3) a=3, L=-6 and \varepsilon=0.005

pls am sorry i can't get the formatting of the question right, kindly help on solving this. thanks!

2. Re: limit of a function!

Originally Posted by lawochekel
find a $\delta$ such that for the given value of L and $\varepsilon$ in the statement $|f(x)-L|\le\varepsilon$ holds whenever $0\le|x-a|\le\delta$
given that $f(x)=(x^2-9)/(x+3)$, $a=3$, $L=-6$ and $\varepsilon=0.005$.
You probably mean a = -3 because f(x) is close to 0, not -6, in the neighborhood of x = 3. Note that f(x) = x - 3 everywhere except for x = -3. I recommend drawing a graph of f(x).

This question belongs in the Pre-Calculus or Calculus forum.