Question

What is `lim_(x->0) (x^3+12x^2-5x)/(5x)` ?

Answer 1

The limit is -1.

Explanation:

`lim_{x->0} (x^3+12x^2-5x)/(5x)=0/0`

this means that we can apply the rule of L'Hôpital and study the derivatives

`lim_{x->0} (x^3+12x^2-5x)/(5x)`

`=lim_{x->0} (d/dx(x^3+12x^2-5x))/(d/dx 5x)`

`=lim_{x->0} (3x^2+24x-5)/5`

`=-5/5=-1`.

Answer 2

`-1`

Explanation:

Note that:

`(x^3+12x^2-5x)/(5x) = 1/5x^2+12/5x-1`

with exclusion `x != 0`

Hence:

`lim_(x->0) (x^3+12x^2-5x)/(5x)= lim_(x->0) (1/5x^2+12/5x-1)`

`= 0+0-1 = -1`

What we have here is a polynomial function with a removable singularity a.k.a. 'hole'. Polynomials are continuous everywhere on their domain.

graph{(y-(x^3+12x^2-5x)/(5x))(x^2+(y+1)^2-0.002) = 0 [-2.656, 2.344, -2.01, 0.49]}