# limit of a function!

• Mar 21st 2012, 03:22 AM
lawochekel
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!
• Mar 21st 2012, 06:04 AM
emakarov
Re: limit of a function!
Quote:

Originally Posted by lawochekel
find a $\displaystyle \delta$ such that for the given value of L and $\displaystyle \varepsilon$ in the statement $\displaystyle |f(x)-L|\le\varepsilon$ holds whenever $\displaystyle 0\le|x-a|\le\delta$
given that $\displaystyle f(x)=(x^2-9)/(x+3)$, $\displaystyle a=3$, $\displaystyle L=-6$ and $\displaystyle \varepsilon=0.005$.

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.