Question

Let f(x) = x^2
2 + 6x. Use the definition (ε − δ method) to show that
lim
x→−3
f(x) = −9 .

Answer 1

We need to prove that for every positive $\epsilon$ > 0, there exist a $\delta$ such that |f(x) + 9| < $\epsilon$ for all x satisfying 0<|x+3|<$\delta\\ |(x^2+6x)+9|<\epsilon\\ |(x+3)(x+3)|<\epsilon\\ \text{for all x satisfying} 0<|x+3|<\delta\\ \text{since |x+3|<} \delta\\ \delta^2<\epsilon\\ \delta<\sqrt\epsilon\\ \text{we choose}\\ \delta=\sqrt\epsilon\\ \text{then the statement}\\ |(x+3)(x+3)|<\epsilon\\ \text{for all x satisfying} 0<|x+3|<\delta\\ \text{holds}\\ Hence\\ Lim_{x\to 2}(x^2+6x)=-9$