Answer to Question #253330 in Calculus for Vistolina

Question

Question 3 Verify using the definition of limit (with ϵ and δ) that lim x→1 (1/2 |x − 1| + 3) = 3, explain all the steps and the overall significance of what you do. [16]

Answer 1

Let us verify using the definition of limit (with $\varepsilon$ and $\delta$ ) that $\lim\limits_{ x→1} (\frac{1}{2} |x − 1| + 3) = 3.$ Let $\varepsilon > 0$ be arbitrary. Put $\delta=\varepsilon.$ Then for each $\varepsilon > 0$ there exists $\delta>0$ such that if $|x-1|<\delta$ then $|(\frac{1}{2} |x − 1| + 3) - 3|=|\frac{1}{2} |x − 1||=\frac{1}{2} |x − 1|<\frac{1}{2}\delta=\frac{1}{2}\varepsilon<\varepsilon.$ Therefore,

$\lim\limits_{ x→1} (\frac{1}{2} |x − 1| + 3) = 3.$