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.