Simple question about using L'Hopital's Rule to solve limits

  • Thread starter Chandasouk
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  • #1
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I know L'Hoptal's rule can only be use when you have indeterminate forms and such, but I do not know how to make a limit a fraction sometimes. Take this limit for example:

The limit at which X approaches infinity of (1+[tex]\frac{1}{x}[/tex])X

I make y = The limit at which X approaches infinity of (1+[tex]\frac{1}{x}[/tex])X

Take the Ln of both sides to get

lny = X*ln(1+[tex]\frac{1}{x}[/tex])

which can be rewritten as ln(1+1/x)/(1/x) but how? I get lost in the algebra and don't know how to obtain that.
 

Answers and Replies

  • #2
Dick
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x=1/(1/x).
 
  • #3
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Multiplying by x is the same as dividing by 1/x (assuming x is not 0).
 
  • #4
HallsofIvy
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I know L'Hoptal's rule can only be use when you have indeterminate forms and such, but I do not know how to make a limit a fraction sometimes. Take this limit for example:

The limit at which X approaches infinity of (1+[tex]\frac{1}{x}[/tex])X

I make y = The limit at which X approaches infinity of (1+[tex]\frac{1}{x}[/tex])X

Take the Ln of both sides to get

lny = X*ln(1+[tex]\frac{1}{x}[/tex])

which can be rewritten as ln(1+1/x)/(1/x) but how? I get lost in the algebra and don't know how to obtain that.
One of the things you learned way back in elementary school was "to divide by a fraction, invert and multiply". Dividing by 1/x is the same inverting to get x and then multiplying:
[itex]x ln(1+ 1/x)= ln(1+ 1/x)/(1/x)[/itex]

This also works the other way: multiplying by a fraction is the same as dividing by its reciprocal: multiplying by x (which is the fraction x/1) is the same as dividing by its reciprocal, 1/x.
 
  • #5
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Holy crap. Thanks guys. It's weird how I forget simple stuff but can do the "calculus" fine.
 

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