# Evaluate the limit, if it exist: lim_{x to infinity} ({(pi x + 1) (3 - x^2)} / {x^3 - pi}).

## Question:

Evaluate the limit, if it exist:

{eq}\displaystyle \lim_{x \to \infty} \bigg(\dfrac {(\pi x + 1) (3 - x^2)} {x^3 - \pi}\bigg) {/eq}.

## Limit:

Suppose that {eq}s(x) {/eq} is a function that is defined on an interval that contains {eq}x=t {/eq}. Then, limit is defined as:

{eq}\mathop {\lim }\limits_{x \to t} s(x) = R {/eq}

For a very small number {eq}b {/eq} such that {eq}b>0 {/eq}, there exists {eq}a>0 {/eq} such that {eq}|s(x) - R| < b {/eq} whenever {eq}0 < |x - t| < a {/eq}.

The following rules are relevant to this problem:

1.{eq}{\frac{r}{r} = 1} {/eq}

2.{eq}{\mathop {\lim }\limits_{r \to \infty } \frac{1}{r} = 0} {/eq}

Given that: {eq}\displaystyle \mathop {\lim }\limits_{x \to \infty } \frac{{(\pi x + 1)(3 - {x^2})}}{{{x^3} - \pi }} {/eq}

{eq}\displaystyle \eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{(\pi x + 1)(3 - {x^2})}}{{{x^3} - \pi }} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}\left( {\pi + \frac{1}{x}} \right)\left( {\frac{3}{{{x^2}}} - 1} \right)}}{{{x^3}\left( {1 - \frac{\pi }{{{x^3}}}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Taking highest term common}}} \right) \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {\pi + \frac{1}{x}} \right)\left( {\frac{3}{{{x^2}}} - 1} \right)}}{{\left( {1 - \frac{\pi }{{{x^3}}}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Cancel the common term}}} \right) \cr & \frac{{\left( {\pi + 0} \right)\left( {0 - 1} \right)}}{{\left( {1 - 0} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\mathop {\lim }\limits_{r \to \infty } \frac{1}{r} = 0} \right) \cr & - \pi \cr & \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{(\pi x + 1)(3 - {x^2})}}{{{x^3} - \pi }} = - \pi \cr} {/eq}