# On finding a limit

So, my homework says:

Find a value of $n \in\mathbb N$ from which there's certainty that:

...

c) $\frac{(-1)^n}{n+1}+2$ is within 1.9 and 2.1

I read this as "make sure that the limit is $2$, within a margin of $2.1 - 1.9$", which, turns into this:

$\left|\frac{(-1)^n}{n+1} +2 \right| < 2.1-1.9$

$(-1)^n < (\frac{1}{5}-2)(n+1)$

$(-1)^n < \frac{-9}{5}n - \frac{9}{5}$

$(-1)^n+\frac{9}{5}n<\frac{-9}{5}$

Now I'm stuck and don't know how to move on: assuming what I deduced is right, how am I supposed to get rid of that $(-1)^n$?

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Hint: We want $$1.9 \lt \frac{(-1)^n}{n+1}+2 \lt 2.1.$$ This is equivalent to $$-0.1\lt \frac{(-1)^n}{n+1}\lt 0.1,$$ since subtracting a constant from both sides of an inequality yields an equivalent inequality.

So we want the absolute value of $\frac{(-1)^n}{n+1}$ to be less than $0.1$. This means that we want $\frac{1}{n+1}$ to be less than $0.1$.

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Your intuition "make sure that the limit is 2, within a margin of 2.1−1.9" is close, but let's clarify a few things. The limit IS 2, meaning that as $n$ gets bigger and bigger, the value of $\frac{(-1)^n}{n+1} + 2$ gets closer and closer to $2$.

In this case, you want to see how big $n$ has to get before you can be sure it is "within a margin of 2.1 - 1.9", or more specifically, make sure that $\frac{1}{n+1}$ is no more than $0.1$ (i.e. you are adding OR subtracting no more than 0.1 from 2. This intuitively "gets rid " of the $(-1)^n$, we can do it rigorously also).

So, how big does $n$ have to be before $\frac{1}{n+1} < 0.1$? If $n>10$, that will work.

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