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

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Answer 1 · 2016-06-07T09:12:01

The limit is -1.

#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 · 2016-07-01T06:25:56

#-1#

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]}