Question

Why does `lim_(x->+-oo) f(x) = lim_(x->0) f(1/x)`?

I know that it is a rule, but I am not sure why.

Answer 1

Suppose you have:

`lim_(x->oo) f(x) = L`

this means that for any number `epsilon >0` we can find a `M_epsilon > 0` such that:

`x > M_epsilon => abs (f(x)-L) < epsilon`

Pose now: `t=1/x`, Clearly when `x > M_epsilon` then `0 < t < 1/M_epsilon`.
So for any `epsilon > 0` we can pose `delta_epsilon = 1/M_epsilon` and have:

`t in (0, delta_epsilon) => abs(f(1/t) - L) < epsilon`

which proves that:

`lim_(x->0^+) f(1/x) = L`

Similarly we can prove that:

`lim_(x->-oo) f(x) = L => lim_(x->0^-) f(1/x) = L`

But if the right and left limits are equal, the function has that value as limit:

`lim_(x->0^+) f(1/x) = lim_(x->0^-) f(1/x) = L => lim_(x->0) f(1/x) = L`