L'Hospital's rule (products)

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  • #1
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Main Question or Discussion Point

To deal with the indeterminate form ##0⋅\pm \infty##, we write the product ##f(x)g(x)## as ##\frac{f(x)}{1/g(x)}## or ##\frac{g(x)}{1/f(x)}##, before applying L'Hospital's rule to one of these forms.
However, on occasion, applying L'Hospital's rule to one of these forms gets us nowhere (##\lim_{x \rightarrow -\infty} x e^x## for instance), despite working for the other (equivalent) quotient.
Is there a way to know beforehand whether ##f(x)g(x)## should be expressed as ##\frac{f(x)}{1/g(x)}## or ##\frac{g(x)}{1/f(x)}##? Or is it always based on trial and error?
 

Answers and Replies

  • #2
RUber
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I think there is some intuition needed. Usually, any function that can be reduced away should be in a position to allow that to happen, as in your example x exp(x).
By leaving the polynomial part on top, you can be sure that application of L'Hopital's rule will eventually reduce that away.
Other than that, I think guess and check until you've done enough...then educated guess and check.
 
  • #3
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To deal with the indeterminate form ##0⋅\pm \infty##, we write the product ##f(x)g(x)## as ##\frac{f(x)}{1/g(x)}## or ##\frac{g(x)}{1/f(x)}##, before applying L'Hospital's rule to one of these forms.
However, on occasion, applying L'Hospital's rule to one of these forms gets us nowhere (##\lim_{x \rightarrow -\infty} x e^x## for instance), despite working for the other (equivalent) quotient.
You can write ##x e^{x}## as ##\frac x {e^{-x}}##. In that form you have the form ##[\frac{-\infty}{\infty}]##, so L'Hopital's Rule can be applied.
MohammedRady97 said:
Is there a way to know beforehand whether ##f(x)g(x)## should be expressed as ##\frac{f(x)}{1/g(x)}## or ##\frac{g(x)}{1/f(x)}##? Or is it always based on trial and error?
 

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