Limit of sequence differ from 2 by 0.02 For what value of $n≥0$, does $(2n+3)/(n+4)$ differ from 2 by 0.02?
I tried to solve this by applying the definition of limit,
$|(2n+3)/(n+4)-2|<0.02$
But didn't get the answer. I think I am misunderstanding the question. Please tell me how to solve this?
Edit:
Answer is $n=196$
This question is from the book engineering mathematics by K A Stroud

If $n\ge 0$ then $n+4>0$ so $|n+4|=n+4,$ and $$\left|\frac {2n+3}{n+4}-2\right|=\left|\frac {2n+3}{n+4}-\frac {2(n+4)}{n+4}\right|=$$ $$=\left|\frac {2n+3-2(n+4)}{n+4}\right|=$$ $$=\left|\frac {2n+3-2n-8}{n+4}\right|=$$ $$=\left|\frac {-5}{n+4}\right|=\frac {|-5|}{|n+4|}=\frac {5}{n+4}.$$ Therefore if $n\ge 0$ then $$\left|\frac {2n+3}{n+4}-2\right|=0.02 \iff $$ $$\iff \frac {5}{n+4}=0.02 \iff$$ $$\iff \frac {n+4}{5}=(0.02)^{-1}=50 \iff $$ $$\iff n+4=(5) \frac {n+4}{5}=(5)(50)=250 \iff$$ $$\iff n=(n+4)-4=250-4=246.$$
This is not about limits. "$X$ and $Y$ differ by $0.02$" means $|X-Y|$ is $exactly$ $0.02.$
If you want to know which $n\ge 0$ satisfy $|\frac {2n+3}{n+4}-2|<0.02$ then from the first part above, we have $$n\ge 0\implies 0<\frac {5}{n+4}=\left|\frac {2n+3}{n+4}-2\right|.$$ So if $n\ge 0$ then  $$\left|\frac {2n+3}{n+4}-2\right|<0.02 \iff $$ $$ \iff 0<\frac {5}{n+4}<0.02 \iff$$ $$ \iff \frac {n+4}{5}>(0.02)^{-1}=50 \iff  n>246.$$
