Question

Question #df591

Answer 1

`lim_(x->0)-sqrt(x) = 0`

Explanation:

The `epsilon-delta` definition of a limit states that given a function `f:D->RR` (a function with the domain `D` and codomain `RR`), then we say that `f` has a limit of `L` at a value `a` if for for every `epsilon > 0` there exists a `delta > 0` such that for any `x in D` where `0 < |x - a| < delta`, we have `|f(x) - L| < epsilon`.

In other words, `lim_(x->a)f(x)=L` if we can make `f(x)` arbitrarily close to `L` by choosing values of `x` from the domain of `f` which are close to `a`.

Using this, we can show that `lim_(x->0)-sqrt(x) = 0`.

Proof: Let `epsilon > 0` be arbitrary. Let `delta = epsilon`. Then, for any `x in [0, oo)`, if `|x - 0| < delta`, we have

`|-sqrt(x) - 0| = |-sqrt(x)| = |sqrt(x)| <= x = |x - 0| < delta = epsilon`.

Thus, by the above definition, with `f(x) = -sqrt(x)`, and `a = L = 0`, we have `lim_(x->0)-sqrt(x) = 0`.

Answer 2

It depends on how you have defined the limit.

Explanation:

For the definition of `lim_(xrarra)f(x)=L`, if we begin by requiring

`f` is defined (and real valued) on some open interval containing `a` except possibly at `x = a`,

then `lim_(xrarr0)(-sqrtx)` is not defined.

For a limit from the right, `lim_(xrarra^+) f(x) = L`, we simply required that `f` is defined on some interval of the form `(a,b)`.

In this case `lim_(xrarr0^+) (-sqrtx)` is defined and is `0`.

If one uses the definition cited by sente in that answer, then, as shown, `lim_(xrarr0) (-sqrtx) = 0`