Question

We know that limit of (3n/(n+4)) when n goes to infinity is 3. If the definition of convergence of sequences, ε=10^(-5) then determine the cutoff index N.

Answer 1

$$\big|\dfrac{3n}{n+4}-3\big|<10^{-5} \text{ whenever}\ n>N$$

$$\big|\dfrac{3n-3n-12}{n+4}\big|<10^{-5} \text{ whenever}\ n>N$$

$$\big|\dfrac{-12}{n+4}\big|<10^{-5} \text{ whenever}\ n>N$$

$$\dfrac{12}{10^{-5}}<n+4\text{ whenever}\ n>N$$

$$n>1200000-4$$

$$N=1199996$$