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

Jun 7, 2016

The limit is -1.

#### Explanation:

${\lim}_{x \to 0} \frac{{x}^{3} + 12 {x}^{2} - 5 x}{5 x} = \frac{0}{0}$

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

${\lim}_{x \to 0} \frac{{x}^{3} + 12 {x}^{2} - 5 x}{5 x}$

$= {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left({x}^{3} + 12 {x}^{2} - 5 x\right)}{\frac{d}{\mathrm{dx}} 5 x}$

$= {\lim}_{x \to 0} \frac{3 {x}^{2} + 24 x - 5}{5}$

$= - \frac{5}{5} = - 1$.

Jul 1, 2016

$- 1$

#### Explanation:

Note that:

$\frac{{x}^{3} + 12 {x}^{2} - 5 x}{5 x} = \frac{1}{5} {x}^{2} + \frac{12}{5} x - 1$

with exclusion $x \ne 0$

Hence:

${\lim}_{x \to 0} \frac{{x}^{3} + 12 {x}^{2} - 5 x}{5 x} = {\lim}_{x \to 0} \left(\frac{1}{5} {x}^{2} + \frac{12}{5} x - 1\right)$

$= 0 + 0 - 1 = - 1$

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]}