Find the smallest value of N that will make the inequality hold for all n N.
-$$a _ { n } = n \tan \frac { 1 } { n }$$
A) $\mathrm { L } = 0 , \mathrm {~N} = 28$
B) $\mathrm { L } = 1 , \mathrm {~N} = 28$
C) $L = 1 , N = 6$
D) diverges

Question

Find the smallest value of N that will make the inequality hold for all n > N.
-$$a _ { n } = n \tan \frac { 1 } { n }$$
A) $\mathrm { L } = 0 , \mathrm {~N} = 28$
B) $\mathrm { L } = 1 , \mathrm {~N} = 28$
C) $L = 1 , N = 6$
D) diverges

Quizplus · Accepted Answer

We have $$\lim _ { n \rightarrow \infty } a _ { n } = \lim _ { n \rightarrow \infty } n \tan \frac { 1 } { n } = \lim _ { n \rightarrow \infty } \frac {\tan \frac { 1 } { n }}{\frac 1n} = \lim _ { x \rightarrow 0 } \frac {\tan x}{x} = 1.$$ Thus, we want to find the smallest $N$ such that for $n>N$, we have $n\tan \frac1n > 1$. Since $\lim_{x\rightarrow0}\frac{\tan x}{x}=1$, we know that there exists a positive real number $\epsilon$ such that for any $01/2$. Choosing $\epsilon = 1/2$, we have $$\frac{\tan(1/2n)}{1/2n} > \frac{1}{2} \Rightarrow n\tan\frac1n > 1$$ whenever $n>1/\epsilon=2$. Thus, we can choose $N=6$ to ensure the inequality holds for all $n>N$.

Find the Smallest Value of N That Will Make the Inequality