Question

Use apsolen delta definition to show that lim x approches to -2 for f(x) 1÷x+1=-1

Answer 1

$lim \, f(x) \text{defined by} \, \dfrac{1}{x+1} = -1 \, \text{as} \, x -> -2 \\ \text{By epsilon-delta definition of limit.}\\ \text{Given,} \, \epsilon > 0, \, \\ \text{we want to find a} \, \delta > 0 \\ \text{such that} \, 0 < |x - x_0| < \delta \implies |f(x) +1| < \epsilon \\ \text{Hence, we have}\\ |\dfrac{1}{x + 1} - (-1)| = |\dfrac{x+2}{x+1}| \\ |\dfrac{1}{x+1}||x+2| \\ \text{Since we are interested in values}\\ \text{closer to 2, we may assume that}\\ |x + 2| < 1 \implies -1 < x + 2 < 1 \\ -3 < x < -1 \implies -3 < x < -1 < 3 \\ -3 < x < 3 \implies -3 + 1 < x + 1 < 3 +1 \\ -2 < x + 1 < 4 \\ \implies \dfrac{1}{4} < \dfrac{1}{x+1} < -\dfrac{1}{2} \\ \dfrac{1}{4} < \dfrac{1}{x+1} < -\dfrac{1}{2} < -\dfrac{1}{4} \\ \implies \dfrac{1}{4} < \dfrac{1}{x+1} < -\dfrac{1}{4} \\ \implies |\dfrac{1}{x+1}| < \dfrac{1}{4} \\ \text{Hence, we see that}\\ |f(x) - (-1)| < |\dfrac{1}{x+1}||x+2| < \epsilon \\ \implies \dfrac{1}{4} |x + 2| < \epsilon \\ \implies |\dfrac{1}{x + 1} - (-1)| < \epsilon \\ \text{provided that} \, |x + 2| < 4 \epsilon \\ \text{Thus, given} \, \epsilon > 0, \\ \text{choose} \, \delta = min{\{1, 4 \epsilon}\}\\ \text{With this choice of} \, \delta \, \\ \text{we have} \, 0 < |x + 2| < \delta \\ \implies |f(x) +1| < \epsilon \\ \text{showing that} \, lim f(x) \, \\ \text{defined by} \\ \dfrac{1}{x+1} = -1 \, \text{as} \, x -> -2$