Following limit is to be evaluated

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.

 

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

 

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

 

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.