Numerator and denominator have both an infinite limit at the given point.

  1. Take $ f(x)=x^2$ and $ g(x)=x^4$ ; then we have

    $\displaystyle \underset{x \rightarrow +\infty}{\lim} f(x)=+\infty$   and$\displaystyle \qquad \underset{x \rightarrow +\infty}{\lim} g(x)=+\infty.$    

    Here we have

    $\displaystyle \frac {f(x)}{g(x)}=\frac{x^2}{x^4}=\frac {1}{x^2},$    

    and

    $\displaystyle \underset{x \rightarrow +\infty}{\lim} \frac{f(x)}{g(x)}=0.$    

  2. Take $ f(x)=x^4$ and $ g(x)=x^3$ ; then we have

    $\displaystyle \underset{x \rightarrow +\infty}{\lim} f(x)=+\infty$   and$\displaystyle \qquad \underset{x \rightarrow +\infty}{\lim} g(x)=+\infty.$    

    Here we have

    $\displaystyle \frac {f(x)}{g(x)}=\frac{x^4}{x^3}=x,$    

    and the function $ f/g$ has a limit at $ +\infty$ , which is equal to $ +\infty$ .

Possible method: Find a common factor in numerator and denominator, which ``pushes'' both towards infinity, and cancel it.

Example 4.4.3   Find the limit at $ +\infty$ of $ f(x)=(2x^3+5x+2)/(x^6+1)$ .

We can easily show that

$\displaystyle \underset{x \rightarrow +\infty}{\lim}(2x^3+5x+2)=+\infty$   and$\displaystyle \qquad \underset{x \rightarrow +\infty}{\lim}(x^6+1)=+\infty,$    

thus we have an undeterminate case. Let us write $ f(x)$ in a more suitable form:

$\displaystyle f(x)=\frac{x^3 \; \left( 2+\frac {5}{x^2} + \frac {2}{x^3} \right...
...{2+\frac {5}{x^2} + \frac {2}{x^3} }{x^3 \; \left( 1 + \frac {1}{x^6} \right)}.$    

As

$\displaystyle \underset{x \rightarrow +\infty}{\lim} \frac {1}{x^2}= \underset{...
...{\lim} \frac {1}{x^3}= \underset{x \rightarrow +\infty}{\lim} \frac {1}{x^6}=0,$    

and

$\displaystyle \underset{x \rightarrow +\infty}{\lim} x^3 = +\infty,$    

we obtain that

$\displaystyle \underset{x \rightarrow +\infty}{\lim} f(x)=0.$    

In fact we saw here an example of what is explained extensively for rational functions in subsection 5.1.

Noah Dana-Picard 2007-12-28