Numerator and denominator have both a limit equal to 0 at the given point.

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

    $\displaystyle \underset{x \rightarrow 0}{\lim} f(x)=0$   and$\displaystyle \qquad \underset{x \rightarrow 0}{\lim} g(x)=0.$    

    Here we have

    $\displaystyle \frac {f(x)}{g(x)}=\frac{x^2}{x^3}=\frac 1x,$    

    and the function $ f/g$ has two different infinite one-sided limits at 0.
  2. Take $ f(x)=x^4$ and $ g(x)=x^3$ ; then we have

    $\displaystyle \underset{x \rightarrow 0}{\lim} f(x)=0$   and$\displaystyle \qquad \underset{x \rightarrow 0}{\lim} g(x)=0.$    

    Here we have

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

    and the function $ f/g$ has a limit at 0, which is equal to 0.

Possible method: Find a common factor in numerator and denominator, which ``pushes'' both towards 0, and cancel it.

Example 4.4.2   Find the limit at 3 of $ f(x)=\frac {x^2-2x-3}{2x^2-5x-3}$ .

First, we compute separately the limit at 3 of the numerator and of tne denominator; we have:

$\displaystyle \underset{x \rightarrow 3}{\lim} (x^2-2x-3) =\underset{x \rightarrow 3}{\lim} (2x^2-5x-3)=0.$    

Therefore, we have here an undeterminate case. Factorize niumerator and denominator:

$\displaystyle \forall x \in \mathbb{R}-\{3,-1/2 \}, \; f(x)=\frac {(x-3)(x+1)}{(x-3)(2x+1)}=\frac{x+1}{2x+1}.$    

It follows:

$\displaystyle \underset{x \rightarrow 3}{\lim} f(x)=\underset{x \rightarrow 3}{\lim} \frac{x+1}{2x+1}=\frac 47.$    

Noah Dana-Picard 2007-12-28