# Find the limit. \lim_{x \rightarrow 0} \frac{x^{3}+12x^{2}-5x}{5x}

## Question:

Find the limit.

{eq}\displaystyle \lim_{x \rightarrow 0} \frac{x^{3} + 12x^{2} - 5x}{5x} {/eq}

## Limit of a Function:

Assume that {eq}f {/eq} is a real valued function of {eq}\displaystyle x {/eq}. Then a real number {eq}l {/eq} is said to be the limit of the function {eq}f {/eq} at a point {eq}a {/eq} in the domain of {eq}f {/eq} if for each {eq}\epsilon>0 {/eq} there exists a {eq}\delta >0 {/eq} such that {eq}\left| f(x) - l \right| < \epsilon {/eq} for all {eq}x {/eq} satisfying {eq}\left| {x - a} \right| < \delta {/eq} and donted by {eq}\mathop {\lim }\limits_{x \to a} f(x) {/eq}.

Here we have to find the limit of {eq}\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 12{x^2} - 5x}}{{5x}}. {/eq}

So we have the following:

{eq}\displaystyle \begin{align} \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 12{x^2} - 5x}}{{5x}}& = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3}}}{{5x}} + \frac{{12{x^2}}}{{5x}} - \frac{{5x}}{{5x}} \hfill \\ &= \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{5} + \frac{{12x}}{5} - 1 \hfill \\ &= \frac{1}{5}\mathop {\lim }\limits_{x \to 0} {x^2} + \frac{{12}}{5}\mathop {\lim }\limits_{x \to 0} x - 1 \hfill \\ &= - 1 \hfill \\ \end{align} {/eq}

Hence we can conclude that {eq}\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 12{x^2} - 5x}}{{5x}} = - 1. {/eq} 