Question

For the given function f(x) and the given values of c and ε>0, find L=lim x→c

f(x) Then determine the largest value for δ>0 such that 0<∣x−c∣<δ→∣f(x)−L∣<ε f(x)= x+3
x 2
+9x+18
,c=−3,ε=0.7 The value of L is (Simplify your answer.)

Answer 1

To answer this question, we need to calculate limit L as x approaches to c of the function f(x), where f(x) = x^2 + 9x + 18, and c = -3.

In the next step, to find L, we substitute '-3' for x in the equation f(x) = x^2 + 9x + 18. Calculating we have:

L = f(c) = (-3)^2 + 9(-3) + 18 = 9 - 27 + 18 = 0.

So we found that L = 0.

Now we need to find δ.

We know that 0 < |x - c| < δ implies |f(x) - L| < ε for p > 0. Here ε = 0.7.
So we have to find δ > 0 that will lead to |x - c| < δ → |f(x) - L| < 0.7.

This means that |(x^2 + 9x + 18) - 0| < 0.7 or |x^2 + 9x + 18| < 0.7.

The inequality |x^2 + 9x + 18| can be solved in two parts:

x^2 + 9x + 18 < 0.7 and -x^2 - 9x -18 < 0.7.
Those inequalities can be solved by keeping x on one side of the inequality and moving other terms to the other side.

From the solutions of those inequalities, δ will be the smallest positive solution.

This ends the calculation of L and δ.