Question

Question #11a56

Answer 1

To find that the limit is `0`, I would reason as follows.

Explanation:

If `x` is very near `5`, then `absx` is very near `abs5` which is `5`.

So, if `x` is very near `5`, then `absx-5` is very near `abs5-5` which is `0`.

Answer 2

As `x` is positive around `x=5` then:

`lim_(x->5) (absx-5) = lim_(x->5) (x-5) = 0`

Explanation:

Consider the limit:

`lim_(x->5) (absx-5) = L`

what we need to prove is that for any `epsilon > 0` we can choose a `delta_epsilon > 0` such that:

`abs(x-5) < delta_epsilon => abs ( abs (x) - 5 -L) < epsilon`

Now if we choose for any `epsilon` a `delta_epsilon < 5`, we have that:

`abs(x-5) < delta_epsilon => 0 < 5-delta_epsilon < x < 5+delta_epsilon`

In other words, `x` is positive and `absx = x`

Thus:

`lim_(x->5) (absx-5) = lim_(x->5) (x-5) = 0`

In general, if we have:

`lim_(x->x_0) f(x)`

to determine the limit we only need to analyze the behavior of the function around `x_0`. So there are three cases:

1) `x_0 > 0`

In this case we can restrict the analysis to an interval where `x>0` and in that interval `abs x = x`

2) `x_0 < 0`

In this case we can restrict the analysis to an interval where `x<0` and in that interval `abs x = -x`

3) `x_0 = 0`

In this case in any interval containing `x_0` there are positive and negative values of `x`, but we know that if the limit exists then:

`lim_(x->0) f(x)`

must coincide with the right and left limit:

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

But `x->0^+` means `x > 0` so `absx =x` and viceversa `x->0^-` means `x < 0` so `absx =-x`

Answer 3

Adding Socratic graph showing the limit `x to +-5_(+-) = 0`.

Explanation:

The other two answers have given details.

The graph here,for `y = |x|-5`, depicts that the limit is 0, either way `x to +-5`..
graph{|x|-5 [-12, 12, -6, 6]}