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Following limit is to be evaluated

  • #1
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the following limit is to be evaluated as x-->infinity

[tex](\frac{x-1}{x+4})^{3x+1}[/tex]

here is the work i've done

taking the natural log:

[tex]3x+1 ln(\frac{x-1}{x+4})[/tex]

to make it an indeterminant form i write it in the following way:

[tex]\frac{ln\frac{x-1}{x+4}}{\frac{1}{3x+1}}[/tex]

applying l'hopital's rule to this yields:

[tex]\frac{\frac{x+4}{x-1}}{\frac{-3}{(3x+1)^2}}[/tex]

simplifying:

[tex]\frac{x+4(3x+1)^2}{-3(x-1)}[/tex]

now noticing that the degree of the numerator is 3 and the denominator is 1, the limit as x--> infinity is infinity.. exponentiating gives e to the infinity....the answer to this is apparently 0 as i found when i graphed the function...where did i go wrong and how to correct this?
 
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Answers and Replies

  • #2
Tom Mattson
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thenewbosco said:
applying l'hopital's rule to this yields:

[tex]\frac{\frac{x+4}{x-1}}{\frac{-3}{(3x+1)^2}}[/tex]
That step is wrong. If you differentiate [itex]\ln(\frac{x-1}{x+4})[/itex], you get:

[tex]\frac{d}{dx}\ln(\frac{x-1}{x+4})=\frac{x+4}{x-1}\frac{d}{dx}(\frac{x+4}{x-1})[/itex].

You forgot about the Chain Rule.
 
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  • #3
Just for thought....are you to evaluate it using the limiting process and l'hopital's rule or are you permitted to just eyeball it? Cuase if you only need to eyeball it, I got a neat pointer for ya. If not, no worries
 
  • #4
lurflurf
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That is much like this limit
[tex]\lim_{x\rightarrow\infty}\left(1+\frac{a}{x}\right)^x=e^a[/tex]
try to make use of this
 
  • #5
dextercioby
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I have a hunch it's something like [itex] e^{-15} [/itex].

Hint

[tex] \frac{x-1}{x+4}=1-\frac{5}{x+4} [/tex]

and the one in post #4.

Daniel.
 

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