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
 A: As correctly noted in the comments, the question assumes that the result $K$ of the whole expression is rounded to two decimal digits. As $n$ increases, $K$ progressively increases and asymptotically tends to $2$. Since the question asks for which $n$ the value of $K$ differs from $2$ by $0.02$, considering the rounding level we must search the minimal value of $n$ for which $1.975\leq K \leq 1.985$.
Writing $K$  as $2-5/(n+4)$, from the lower bound we get
$$2-5/(n+4)\geq 1.975$$
and then
$$5/(n+4)\leq 0.025=\frac{1}{40}$$
$$5 \leq \frac{n}{40}  + \frac{1}{10}$$
$$n \geq 40\left(5- \frac{1}{10}\right)=196$$
A: I believe the question should read "For what values of $n\geq 0$ does $\frac{2n+3}{n+4}$ differ from $2$ by at most 0.02".
So you have two inequalities:
$$\frac{2n+3}{n+4} - 2 < 0.02$$
and
$$\frac{2n+3}{n+4} - 2 > -0.02.$$
The task is to solve these two inequalities, i.e. find the values of $n$ for which they both hold.
A: 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  $$\left|\frac {2n+3}{n+4}-2\right|<0.02 \iff $$ $$ \iff 0<\frac {5}{n+4}<0.02 \iff$$ $$ \iff \frac {n+4}{5}>(0.02)^{-1}=50 \iff  n>246.$$
