I was reading a book on introduction to calculus which stated the definition of a limit as follows:
A number '$a$' is said to be the limit of a sequence $y_n$ if for any $\epsilon > 0$, there exists a number $N$ such that for all $n>N$, $|(y_n-a)| < \epsilon$
Then it asked us to prove the following:
Prove that: $\lim_{x\to\infty} \frac{x}{x+1} = 1$
It's a very easy problem and can be easily solved by writing the fraction as $\frac{x+1}{x+1} + \frac{-1}{x+1}$.
Now let us try to evaluate $\lim_{x\to\infty} \frac{x}{x+1}$ without knowing it's answer using first principles ( the definition ).
I had a hard time trying to solve it. This is what I did:
First of all, assume the limit of the sequence is '$a$'. Then find out $|(y_n-a)|$. It comes out to be: $\frac{(an+1)-n}{n+1}$ ( assuming $a$ is $ \geq$ 1 ).
Now according to the definition: $\frac{(an+1)-n}{n+1} < \epsilon$.
At this point, all we know is that $\epsilon$ should be quite small to make our result count and $n$➝$\infty$. How can we find '$a$' using this information?
Let's give it a run: First, you find the limit itself. By plugging $x = 1000; 1,000,000; 1,000,000,000$, the values of $\dfrac{x}{x+1}$ are so close to $1$. Thus your claimed value of the limit is $1$. And $\left|\dfrac{x}{x+1}-1\right|= \dfrac{1}{|x+1|}$. Thus you begin with letting $\epsilon > 0$ be given, you find a $N$ such that if $x > N$, then $\dfrac{1}{|x+1|} < \epsilon$. This translates the problem into some thing you are familiar with: solve $\dfrac{1}{x+1} < \epsilon$. Thus $x > \dfrac{1}{\epsilon} - 1$. Thus you can take $N = \lfloor{\dfrac{1}{\epsilon} - 1}\rfloor+1$. You are done.