# Following limit is to be evaluated

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:

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Tom Mattson
Staff Emeritus
Gold Member
thenewbosco said:
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:

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

dextercioby
Homework Helper
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.