To estimate the value of a limit, let’s say lim f(x) as x->c you guess what f(x) gets really really close too as x gets closer to c. For example let’s say we want to estimate the limit of f(x) = (x^2-1)/(x-1) as x->1. Let’s first notice that f(x) is actually undefined at 1, since the denominator would be 0. But we can still talk about what happens as we get close, so let’s pick some trial numbers that progressively get closer to 1. how about 2,1.5,1.1, 1.01, and 1.0001.
So
f(2) = 3
f(1.5) = 2.5
f(1.1) = 2.1
f(1.01) = 2.01
f(1.0001) = 2.0001
It seems as x gets closer to 1, f(x) gets closer to 2. If we do the same thing from the other side (x values less than 1) it will get close to 2 also. So a good estimation of our limit would be 2. But notice this is by no means an exact or even good process, soon you will learn a formal way to tell exactly what limits are equal too.
For your next question. When does a limit exist? Since I don’t think you have the formal definition of a limit yet I’ll give you an intuitive one. The limit of f(x) as x-> c exist if as x gets arbitrarily close to c then f(x) gets arbitrarily close to some unique value. So let’s take f(x) = 1/x as x-> 0. This limit doesn’t exist because f(x) at 0 doesn’t get close to a value let alone a unique value. In this case we say f(x) diverges at c. Let’s consider another case f(x) = x+1 if x =< 2, x^2 if x > 2. Lim f(x) as x->2 doesn’t exist. Because x+1 gets close to 3, and x^2 gets close to 4. So as x gets close to 2, f(x) doesn’t approach a single value, it approaches 2.
