Homework Help: Following limit is to be evaluated

thenewbosco · Sep 22, 2005

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?

Tom Mattson · Sep 22, 2005

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.

DaMastaofFisix · Sep 22, 2005

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

lurflurf · Sep 23, 2005

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

dextercioby · Sep 23, 2005

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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Homework Help: Following limit is to be evaluated

thenewbosco

Tom Mattson

DaMastaofFisix

lurflurf

dextercioby