Difference between approaching and being exactly a number

Let's look at your sentence

It never reaches $0$, but becomes closer and closer to $0$.

This is imprecise, and this is at the center of your confusion. What is "it"? There is an important distinction to be made:

• The sequence $1,\frac{1}{2},\frac{1}{3},\ldots$ never reaches $0$, but becomes closer and closer to $0$.

• Let $L$ be the limit of the sequence $1,\frac{1}{2},\frac{1}{3},\ldots$. Then $L$ is exactly equal to $0$.

The point is that the limit is exactly the operation that takes a sequence $x_n$ approaching $x$ and gives you $x$. That is, when we say $$\lim_{n \to \infty} x_n = x,$$ we mean that $x_n$ converges to $x$ as $n\to \infty$; this does not claim that $x_n = x$ for any $n$. In your sum, the ellipsis $\cdots$ implies a limiting process: this equation could be more formally written

$$ 1 = \lim_{N \to \infty} \sum_{n=1}^N \frac 1 {2^n}.$$

Note that none of these finite sums is equal to $1$, but they approach $1$, so we say their limit is 1.

Imagine that we say $\forall \epsilon>0 :|x-y|<\epsilon$. I want to claim that if this is true for any arbitrary positive $\epsilon$ then it forces $|x-y|$ to be zero.

Suppose that $|x-y|$ is not zero but it is very very small, so, there exists a natural number $n$ such that

$$\exists n\in\mathbb{N}:\displaystyle \frac{1}{10^{-(n+1)}} <|x-y|< \frac{1}{10^{-n}}$$

Now we can set $\displaystyle \epsilon<\frac{1}{10^{-(n+1)}}$ and that contradicts $|x-y|<\epsilon$. So, because $|x-y|<\epsilon$ is true for any $\epsilon>0$, no matter how small $\epsilon$ is we know $|x-y|=0 \implies x-y=0 \implies x=y$. This is because there is no smallest positive real number. And because $|x-y|$ is non-negative, then if it's not zero, we'll have a contradiction, so it must be zero.

This is true for limits too. When you say $\displaystyle 2=1+\frac{1}{2}+\frac{1}{4}+\cdots$ this is an equality. But if you consider only a finitely many terms of this series then that is only an approximation of $2$.

when you are approaching a number(a->5), it means that you are hypothetically so close to the number, that it is practically possible to treat it =5. derivative means a small change in x, with respect to a small change in y.

since y=sin(x). and we know that slope(derivative in nothing but finding the slope) =(y2-y1)/(x2-x1),therefore dy/dx=(sin(x+small change in x)-sin(x))/small change in x