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Indeterminate forms:

$``\infty - \infty"$  \hfil
$``\infty \cdot 0"$   \hfil
$``{\infty \over \infty}"$	\hfil
$``{0 \over 0}"$

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\centerline{\hfil $``{\infty^0}"$ \hfil $``{0^0}"$ \hfil $``{1^\infty}"$ \hfil}

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L'Hospital's Rule:  Suppose $f'$ and $g'$ exist, $g'(x) \not = 0$ near $a$ except possibly at $a$ where $a \in {\cal R} \cap \{+\infty, -\infty\}$ and if either

1.)  $lim_{x \rightarrow a} f(x) = lim_{x \rightarrow a} g(x) = 0$ 

or

2.)  $lim_{x \rightarrow a} f(x) = \pm \infty$ and $lim_{x \rightarrow a} g(x) = \pm \infty$ 

and if $lim_{x \rightarrow a} {f'(x) \over g'(x)}$ exists, then
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\centerline{$lim_{x \rightarrow a} {f(x) \over g(x)} = lim_{x \rightarrow a} {f'(x) \over g'(x)}$ } 
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Ex 1)  $lim_{x \rightarrow +\infty} {-2x^2 + 3 \over 5x^2 + 4x}$

Old method:  $lim_{x \rightarrow +\infty} {-2x^2 + 3 \over 5x^2 + 4x}$
= $lim_{x \rightarrow +\infty} {x^2(-2 + {3 \over x^2}) \over x^2(5 + {4 \over x})} =$

New method:

$lim_{x \rightarrow +\infty} {-2x^2 + 3 \over 5x^2 + 4x}$

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$lim_{x \rightarrow 1} {1 - x + ln(x) \over x^3 - 3x + 2}$

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