Thread: Tricky limit without using l'Hopital's rule

1. Tricky limit without using l'Hopital's 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?

2. $e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1 }$

Is this what you're trying to verify? So basically you're trying to show this is equal to e by transforming it into $\left(1+\frac{1}{u}\right)^u$? You've been asked to do this without using logarithms or L-hopitals rule?

3. Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:

$e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1 }=\lim_{x->\infty}\left(1+4x)\log_{e}\left(1+\frac{1+x}{4x^2 +4x+4}\right)$

But then we have again a case of $0*\infty$, 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?

4. Originally Posted by soulforged
Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:

$e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1 }=\lim_{x->\infty}\left(1+4x)\log_{e}\left(1+\frac{1+x}{4x^2 +4x+4}\right)$

But then we have again a case of $0*\infty$, 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?
Quick question: are you allowed to assume that $\lim\limits_{x\to\infty}\left(1+\frac{1}{x}\right) ^x=e$? If so, then your limit can be computed in several steps.

5. Originally Posted by soulforged
Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:

$e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1 }=\lim_{x->\infty}\left(1+4x)\log_{e}\left(1+\frac{1+x}{4x^2 +4x+4}\right)$

But then we have again a case of $0*\infty$, 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?
L'Hospital's Rule is often the most elegant path...

6. Originally Posted by Chris L T521
Quick question: are you allowed to assume that $\lim\limits_{x\to\infty}\left(1+\frac{1}{x}\right) ^x=e$? If so, then your limit can be computed in several steps.
Yep, I'm allowed to do that, and this is actually just what I'm looking for!

Could you give a hint or two on how to solve it using this fact?

$e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1 }$

Is this what you're trying to verify? So basically you're trying to show this is equal to e by transforming it into $\left(1+\frac{1}{u}\right)^u$? You've been asked to do this without using logarithms or L-hopitals rule?
$\displaystyle\frac{x+1}{4x^2+4x+4}=\frac{1}{4x+\fr ac{4}{x+1}}$

$\displaystyle\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}=\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]$

As $x\rightarrow\infty$ the 2nd factor $\rightarrow\ 1$

and

$\displaystyle\ 4x+\frac{4}{x+1}\rightarrow\ 4x$

8. 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.

9. If we write for $x>3$

$\displaystyle\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}>\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+\frac{4}{x+1}}>\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x}$

but

$\displaystyle\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}=\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x}$

therefore

$\displaystyle\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}=\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+\frac{4}{x+1}}$

10. Alright then, I get it now. Thank you very much for this explanation.