# What is the limit of [3 + 4/x - 5/x^2 + [x-1]/[x^3+1] as x goes to infinity?

Oct 24, 2015

3

#### Explanation:

Every term in the expression $3 + \frac{4}{x} - \frac{5}{{x}^{2}} + \frac{x - 1}{{x}^{3} + 1}$ goes to zero as $x \rightarrow \infty$ except the 3.

The reason is that the other terms are rational functions where the denominator has a higher degree than the numerator.

Jul 27, 2016

If we actually took the limit, we can separate each term. Since each of these functions have existent limits...

$\textcolor{b l u e}{{\lim}_{x \to \infty} 3 + \frac{4}{x} - \frac{5}{x} ^ 2 + \frac{x - 1}{{x}^{3} + 1}}$

$= {\lim}_{x \to \infty} 3 + {\lim}_{x \to \infty} \frac{4}{x} - {\lim}_{x \to \infty} \frac{5}{x} ^ 2 + {\lim}_{x \to \infty} \frac{x - 1}{{x}^{3} + 1}$

As $x \to \infty$ for $\frac{x - 1}{{x}^{3} + 1}$, the $- 1$ and $+ 1$ become insignificant, so this is equivalent to ${\lim}_{x \to \infty} \frac{1}{{x}^{2}}$:

$\implies 3 + {\cancel{\frac{4}{\infty}}}^{0} - {\cancel{\frac{5}{\infty}}}^{0} + {\cancel{{\lim}_{x \to \infty} \frac{1}{{x}^{2}}}}^{\frac{1}{\infty} \to 0}$

$= \textcolor{b l u e}{3}$