How to calculate $\lim_{x\to\infty} (\frac{x}{x+1})^x$ using L'Hopitals rule?

I am trying to calculate $$\lim_{x\to\infty} (\frac{x}{x+1})^x$$ using L'Hopital.

Apparently without L'Hopital the limit is

$$\lim_{x\to\infty} (\frac{x}{x+1})^x = \lim_{x\to\infty} (1 + \frac{-1}{x+1})^x = \lim_{x\to\infty} (1 - \frac{1}{x+1})^{x+1} \frac{1}{1-\frac{1}{x+1}} = e^{-1} * \frac{1}{1} = \frac{1}{e}$$

I am wondering how one could calculate this limit using L'Hopital's rule.

My failed approach

My initial syllogism was to use the explonential-log trick in combination with the chain rule as following:

$$\lim_{x\to\infty} (\frac{x}{x+1})^x = e^{\lim_{x\to\infty} x \ln(\frac{x}{x+1})} \quad (1)$$

So, basically the problem that way is reduced to:

$$\lim_{x\to\infty} x \ln(\frac{x}{x+1}) = \lim_{x\to\infty} x * \lim_{x\to\infty}\ln(\frac{x}{x+1}) \quad (2)$$

As far as $$\ln(\frac{x}{x+1})$$ is concerned, it has the form $$f(g(x))$$, so using the chain rule for limits and chain rule for derivatives in order to apply L'Hopital we can rewrite it as:

$$\lim_{x\to\infty} \ln( \lim_{x\to\infty} \frac{(x)'}{(x+1)'}) = \lim_{x\to\infty} ln(1) \quad (3)$$

But $$(2),(3) \to 0 * \infty$$, so that failed.

Any ideas on how we could approach this in other ways?

Caution,

$$\lim fg=\lim f\lim g$$ can only be used when the limits on the right both exist, which is not the case here.

By L'Hospital

$$\lim_{x\to\infty}\log\left(\frac x{x+1}\right)^x=\lim_{x\to\infty}\frac{\log\left(\dfrac x{x+1}\right)}{\dfrac1x}=\lim_{x\to\infty}\frac{\dfrac1x-\dfrac1{x+1}}{-\dfrac1{x^2}}=-\lim_{x\to\infty}\frac x{x+1}=-1.$$

The simplest is, by continuity of the inverse function,

$$\lim_{x\to\infty}\left(\frac x{x+1}\right)^x=\frac1{\lim_{x\to\infty}\left(1+\dfrac1x\right)^x}=\frac1e.$$

Note that $$\displaystyle \left( \dfrac{x}{1+x}\right)^x = \left( 1 - \frac{1}{x} \right)^x,$$ so the limit $$L$$ that you're asking is the same as

$$L = \displaystyle \lim_{x \to \infty} \left( 1 - \frac{1}{x}\right)^x$$

Taking the substitution $$u = -1/x$$ we have that

$$L = \displaystyle \lim_{u \to 0^-} \left( 1 + u \right)^{-\frac{1}{u}}$$

Finally, by definition $$e = \displaystyle \lim_{u \to 0} (1 + u)^{1/u}$$ so your limit is $$L = e^{-1}$$.

• very nice that it does not use l'hopital Commented Jan 2 at 23:08

You wrote:

"$$\lim_{x\to\infty} x \ln(\frac{x}{x+1}) = \lim_{x\to\infty} x * \lim_{x\to\infty}\ln(\frac{x}{x+1}).$$"

Observe that $$\lim_{x\to\infty} x = \infty$$ !

For $$\lim_{x\to\infty} x \ln(\frac{x}{x+1})$$ let $$f(x):= \ln(\frac{x}{x+1})$$ and $$g(x):=1/x.$$ Then we have

$$\lim_{x\to\infty} x \ln(\frac{x}{x+1})= \lim_{x\to\infty} \frac{f(x)}{g(x)}.$$

This is a limit of the form "$$\frac{0}{0}$$".

Now compute $$\lim_{x\to\infty} \frac{f'(x)}{g'(x)}.$$

$$x \ln (\frac x {x+1})=\frac {\ln x -\ln (x+1)} {1/x}$$. So we get $$\lim \frac {1/x-1/(1+x)} {-1/x^{2}}=-\lim \frac x {1+x}=-1$$. Hence the answer is $$e^{-1}$$.