Following limit is to be evaluated

[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?

 

[quantumdude]

applying l'hopital's rule to this yields:

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

That step is wrong. If you differentiate $\ln(\frac{x-1}{x+4})$, 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].<br /> <br /> You forgot about the Chain Rule.[/tex]

 

[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

 

[lurflurf]

That is much like this limit
$$\lim_{x\rightarrow\infty}\left(1+\frac{a}{x}\right)^x=e^a$$
try to make use of this

 

[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.