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

##### 2 Answers

The limit is -1.

#### Explanation:

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

Jul 1, 2016

#### Explanation:

Note that:

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

with exclusion

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]}