Question

Evaluating Limits. Does the limit of `x ->-5` exist in the function `(x^3)/(x+5)^2`?

Evaluating Limits. Does the limit of `x ->-5` exist in the function `(x^3)/(x+5)^2`?

Answer 1

`lim_(x->-5) x^3/(x-5)^2 =-oo`

Explanation:

The numerator is continuous for `x=-5` as:

`[x^3]_(x=-5) = -125`

The denominator instead vanishes as:

`lim_(x->-5) (x+5)^2 = 0`

and we can note that it approaches `0` only with positive values as:

`(x+5)^2 > 0` for `x!= -5`

Around `x=-5` then the ratio:

`x^3/(x+5)^2 < 0`

is negative and not bounded, so we can conclude that:

`lim_(x->-5) x^3/(x-5)^2 =-oo`

We can formally demonstrate is by choosing any number `M >0` and then a number `delta_M > 0` such that `delta_M < min(4, 1/sqrt(M))`.

Then, for `x in (-5-delta_M, -5+delta_M)` we have that, as `x^3` is strictly increasing:

`x^3 <= (-5+delta_M)^3`

and then as `delta_M < 4`

`x^3 <= (-5+4)^3`

that is:

`(1) " "x^3 <= -1`

while as:

`abs (x+5) < delta_M`

`(x+5)^2 < delta_M^2`

and as `delta_M < sqrt(1/M)`:

`(x+5)^2 < (sqrt(1/M))^2`

that is:

`(x+5)^2 < 1/M`

and taking the reciprocal:

`(2) " " 1/(x+5)^2 > M`

Multiplying the inequalities `(1)` and `(2)` we finally have that `x in (-5-delta_M, -5+delta_M)` implies

`x^3/(x+5)^2 < -M`

which proves that:

`lim_(x->-5) x^3/(x-5)^2 =-oo`