Following limit is to be evaluated

1. Sep 22, 2005

thenewbosco

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

$$(\frac{x-1}{x+4})^{3x+1}$$

here is the work i've done

taking the natural log:

$$3x+1 ln(\frac{x-1}{x+4})$$

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

$$\frac{ln\frac{x-1}{x+4}}{\frac{1}{3x+1}}$$

applying l'hopital's rule to this yields:

$$\frac{\frac{x+4}{x-1}}{\frac{-3}{(3x+1)^2}}$$

simplifying:

$$\frac{x+4(3x+1)^2}{-3(x-1)}$$

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?

Last edited: Sep 22, 2005
2. Sep 22, 2005

Tom Mattson

Staff Emeritus
That step is wrong. If you differentiate $\ln(\frac{x-1}{x+4})$, you get:

$$\frac{d}{dx}\ln(\frac{x-1}{x+4})=\frac{x+4}{x-1}\frac{d}{dx}(\frac{x+4}{x-1})[/itex]. You forgot about the Chain Rule. Last edited: Sep 22, 2005 3. Sep 22, 2005 DaMastaofFisix Just for thought....are you to evaluate it using the limiting process and l'hopital's rule or are you permitted to just eyeball it? Cuase if you only need to eyeball it, I got a neat pointer for ya. If not, no worries 4. Sep 23, 2005 lurflurf That is much like this limit [tex]\lim_{x\rightarrow\infty}\left(1+\frac{a}{x}\right)^x=e^a$$
try to make use of this

5. Sep 23, 2005

dextercioby

I have a hunch it's something like $e^{-15}$.

Hint

$$\frac{x-1}{x+4}=1-\frac{5}{x+4}$$

and the one in post #4.

Daniel.