Answer to Question #261859 in Calculus for Pankaj

Question

Using Epsilon and Delta definition, show that limit of x approches to 2 for 3x-5 is equal to 1.

Answer 1

Using Epsilon and Delta definition, let us show that $\lim\limits_{x\to 2}3x-5 =1.$

Let $\varepsilon>0$ be arbitrary. Put $\delta=\frac{1}3\varepsilon.$ Then for any $\varepsilon>0$ there exists $\delta>0$ such that if $|x-2|<\delta$ then $|(3x-5)-1|=|3x-6|=3|x-2|<3\delta=3\frac{1}3\varepsilon=\varepsilon.$ We conclude that $\lim\limits_{x\to 2}3x-5 =1.$