What is the limit of ((x+1)/x)^x as x approaches oo?

Jul 21, 2016

${\lim}_{x \rightarrow \infty} {\left(\frac{x + 1}{x}\right)}^{x} = e$

Explanation:

Ok this requires a few wee tricks. We want to find

${\lim}_{x \rightarrow \infty} {\left(\frac{x + 1}{x}\right)}^{x}$

Because the exponential and natural log functions are inverse to each other they cancel out so we can rewrite this as

${\lim}_{x \rightarrow \infty} \exp \left(\ln {\left(\frac{x + 1}{x}\right)}^{x}\right)$

Using rules of logs we can bring the exponent down:

${\lim}_{x \rightarrow \infty} \exp \left(x \ln \left(\frac{x + 1}{x}\right)\right)$

Now notice that the bit that actually changes is the exponent of the exponential function, so that's what we focus on:

$\exp \left({\lim}_{x \rightarrow \infty} \left(x \ln \left(\frac{x + 1}{x}\right)\right)\right)$

If at all possible we want to use L'hopital's rule on the limit. For this we need it to be in indeterminate form, eg $\frac{0}{0} \mathmr{and} \frac{\infty}{\infty} e t c . .$

Rewrite as:

$\exp \left({\lim}_{x \rightarrow \infty} \left(\ln \frac{\frac{x + 1}{x}}{\frac{1}{x}}\right)\right)$

Check the numerator and denominator separately:

${\lim}_{x \rightarrow \infty} \frac{1}{x} = 0$

${\lim}_{x \rightarrow \infty} \ln \left(\frac{x + 1}{x}\right) = {\lim}_{x \rightarrow \infty} \ln \left(\frac{1 + \frac{1}{x}}{1}\right) = \ln \left(1\right) = 0$

so we have $\frac{0}{0}$ and are in the correct form for L'hopital's.

$\exp \left({\lim}_{x \rightarrow \infty} \left(\frac{\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x + 1}{x}\right)\right)}{\frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)}\right)\right)$

Computing the derivative of the numerator using chain and quotient rules:

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x + 1}{x}\right)\right) = \frac{x}{x + 1} \cdot \frac{- 1}{{x}^{2}} = - \frac{1}{x \left(x + 1\right)}$

For the denominator, easiest just to rewrite and use power rule

$\frac{d}{\mathrm{dx}} \left({x}^{- 1}\right) = - \frac{1}{x} ^ 2$

We now have:

$\exp \left({\lim}_{x \rightarrow \infty} \left(\frac{- \frac{1}{x \left(x + 1\right)}}{- \frac{1}{x} ^ 2}\right)\right) = \exp \left({\lim}_{x \rightarrow \infty} \left(\frac{x}{x + 1}\right)\right)$

${\lim}_{x \rightarrow \infty} \left(\frac{x}{x + 1}\right) = {\lim}_{x \rightarrow \infty} \left(\frac{1}{1 + \frac{1}{x}}\right) = 1$

So we end up with

${\lim}_{x \rightarrow \infty} {\left(\frac{x + 1}{x}\right)}^{x} = {e}^{1} = e$