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# Limits

### Limits involving infinity

• If $$n$$ is a positive number, then,
$\lim_{x\rightarrow \infty} \frac{1}{x^n} =0$
• For example:
$\lim_{x\rightarrow \infty} \frac{x^3+5x}{2x^3-x+4}$
If we consider the expression as $$x$$ approaches infinity: both the numerator and the denominator approach infinity, so we do not know what the limit is.
• Firstly, divide each term by the highest power of $$x$$ in the denominator giving:
\begin{eqnarray*}
\lim_{x\rightarrow \infty} \frac{x^3+5x}{2x^3-x+4}
&=& \lim_{x\rightarrow \infty}\displaystyle \frac{\frac{x^3+5x}{x^3}}{\frac{2x^3-x+4}{x^3}} \\
&=& \lim_{x\rightarrow \infty} \frac{1+\frac{5}{x^2}}{2-\frac{1}{x^2}+\frac{4}{x^3}} \\
&=& \frac{\displaystyle \lim_{x\rightarrow \infty} 1 + \lim_{x\rightarrow \infty} \frac{5}{x^2}}{\displaystyle \lim_{x\rightarrow \infty} 2 - \lim_{x\rightarrow \infty} \frac{1}{x^2} + \lim_{x\rightarrow \infty} \frac{4}{x^3} } \\
&=& \frac{1+0}{2-0+0} \\
&=& \frac{1}{2}
\end{eqnarray*}
• Therefore, the limit of the expression, as $$x$$ approaches infinity, is a half.