The famous epsilon-delta definition for finding the limit. The objective is to prove that $\displaystyle \lim_{x \to 2}\dfrac{x^3-4}{x^2+1}=\dfrac{4}{5}$.
For this, we start as follows:
Say for $\epsilon>0$ there exists a $\delta>0$ such that $0<|x-2|<\delta$ and then prove that $|f(x)-\frac{4}{5}|<\epsilon$ where $f(x)=\dfrac{x^3-4}{x^2+1}$.
So, the objective becomes to find this $\delta$ as a function of $\epsilon$ i.e $\delta (\epsilon)$ is required.
Now, the text I am reading does the following,
It lets $\delta=1$ initially and tries to get an upper bound on $\left|f(x)-\dfrac{4}{5}\right|$.
My doubt is, can we really do that ? Our objective is to find such a $\delta$ and we are starting of the proof with an arbitrary value of $\delta$ ? We could have taken $\delta=0.5,0.6,0.7$ or any other positive value and the solution would change accordingly, it's like we want to find a solution for a variable and we find that by assuming an arbitrary value of that variable initially, why ?
I mean something is wrong, may be my interpretation of the definition of the limit is wrong ? Can anyone help ?
 A: The issue is easier to see in something like $\lim_{x \to 1} x^2=1$. You can write $x^2-1=(x+1)(x-1)$. The second factor can be made small by making $\delta$ small. But the first factor can be large potentially. So you make the first factor not a problem by deciding in advance that $\delta$ is less than some constant. For instance if $|x-1|<1$ then $|x+1|<3$, so that $|x^2-1|<3|x-1|$. So $\delta=\min \{ \epsilon/3,1 \}$ works. Note that if you took $\epsilon=6$ then $\delta=\epsilon/3=2$ would fail: $|3^2-1|=8>6$. This is because of that $x+1$ factor getting too big. But $\delta=1$ still works, because $|2^2-1|=3<6$.
This is fine because you are just choosing any function $\delta(\epsilon)$ you want that has the property. It does not have to be as large as possible, nor does it need to be an increasing function. Geometrically this is because continuity at a point is a local property, in that only function values arbitrarily close to the point matter for it.
A: hint: $|f(x) - 4/5| = \dfrac{|5x^3-4x^2-24|}{5(x^2+1)}= \dfrac{|x-2||5x^2+6x+12|}{5(x^2+1)}$. Next use $x^2+1 > 1$, $5x^2+6x+12 \le 5|x|^2+6|x|+12< 5\cdot 3^2+6\cdot 3+12 = 75$. This comes from $|x| \le |x-2|+2 < 1+2 = 3$ for $|x-2| < \delta = 1$ as you chose earlier. Can you put these together to make a proof? You choose $\delta = 1$ for convenience, and you can choose any positive delta small enough to make $|f(x) - 4/5| < \epsilon$.
