Question

How do you find the Limit of `x^2+1` as `x->1` and then use the epsilon delta definition to prove that the limit is L?

Answer 1

The function `f(x) = x^2+1` is a polynomial and as such is defined and continuous for any `x in RR`, so that:

`lim_(x->1) x^2+1 = f(1) = 2`

Now consider any number `epsilon > 0` and evaluate the difference:

`abs (f(x) -2) = abs( x^2 +1 -2) = abs( x^2-1) = abs (x-1) abs (x+1)`

If we choose `delta_epsilon < min (1, epsilon/3)` then we have that:

`x in (1-delta_epsilon, 1+delta_epsilon) => abs(x-1) < delta_epsilon < epsilon/3`

and

`abs(x+1) < abs(2+delta_epsilon) < 3`

so that:

`x in (1-delta_epsilon, 1+delta_epsilon) => abs (f(x) -2) < 3 epsilon/3 = epsilon`