Is this what you're trying to verify? So basically you're trying to show this is equal to e by transforming it into ? You've been asked to do this without using logarithms or L-hopitals rule?
It goes like this:
lim[x->infinity](1 + (x + 1)/(4x^2 + 4x + 4))^(4x+1).
The solution is e, that I got with Mathematica and by myself, using a few applications of l'Hopital's rule, but the calculation is very long and tedious...
Is there any other way to solve it? Could we use the known equality lim[n->infinity](1+1/n)^n = e?
Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:
But then we have again a case of , and another application of L'Hopital's rule is needed, and then another and... then, some derivatives need to be calculated and finally, when all is done, I get the right result e.
I have a hunch there exists a more elegant path to this result, but unfortunately I don't see it at the moment. Care to help?
I guess this makes the problem solved. Thanks.
Although I must admit I'm not entirely comfortable with this solution, because you've jumped from lim(f(x)^g(x)) to lim(f(x))^g(x), and I'm not sure if you can actually do that.
In other words, in the last step, you've calculated the inner limit of (4x + 4/x+1) = 4x, and left the outer expression unafected, but we have an exponentiation there...
I know one can do e.g. lim(f(x)*g(x)) = lim(f(x)) * lim(g(x)), but I know of no theorem that says something along similar lines for the exponentiation... if you see what I'm getting at.