Find the limit of x^(x+1)/(1+x)^x - 1/e as x tends to positive infinity.

Step-by-step solution

1. 1

   Rewrite the quotient so the classic $e$ limit appears. Split off one factor of $x$ from the numerator: $\frac{x^{x+1}}{(1+x)^{x}}=x\cdot\frac{x^{x}}{(1+x)^{x}}=x\left(\frac{x}{1+x}\right)^{x}=x\left(1+\frac{1}{x}\right)^{-x}.$ This is the move that turns an intimidating tower of exponents into something with a known limit inside it.

2. 2

   Evaluate the bracketed factor. The standard limit $\displaystyle\lim_{x\to+\infty}\left(1+\tfrac1x\right)^{x}=e$ gives $\left(1+\frac{1}{x}\right)^{-x}\longrightarrow\frac{1}{e},$ so the factor settles down to the constant $e^{-1}\approx 0.3678794$ - it neither vanishes nor blows up.

3. 3

   Multiply the pieces back together. A factor tending to the positive constant $1/e$ times a factor $x\to+\infty$ diverges: $\frac{x^{x+1}}{(1+x)^{x}}=x\left(1+\frac1x\right)^{-x}\sim\frac{x}{e}\longrightarrow +\infty.$ So the first term of the expression grows without bound, at the linear rate $x/e$.

4. 4

   Subtract the constant. Since $\dfrac1e$ is a fixed finite number, removing it changes nothing about divergence: $\lim_{x\to+\infty}\left(\frac{x^{x+1}}{(1+x)^{x}}-\frac{1}{e}\right)=+\infty.$ A constant can never cancel a divergent term - only something that itself grows like $x/e$ could.

5. 5

   See what a sharper subtraction would give. Expanding $x\ln\!\left(1+\tfrac1x\right)=1-\tfrac{1}{2x}+\tfrac{1}{3x^{2}}-\cdots$ and exponentiating, $x\left(1+\frac1x\right)^{-x}=\frac{x}{e}+\frac{1}{2e}-\frac{5}{24e\,x}+O\!\left(x^{-2}\right).$ Hence subtracting $\dfrac{x}{e}$ instead of $\dfrac{1}{e}$ gives a finite limit, $\lim_{x\to+\infty}\left(\frac{x^{x+1}}{(1+x)^{x}}-\frac{x}{e}\right)=\frac{1}{2e}=0.1839397206,$ which is the version of this problem that textbooks usually intend.

6. 6

   Confirm both statements numerically. At $x=10^{5}$ the expression equals $36788.1280560984$, while $x/e=36787.9441171442$ and $1/e=0.3678794412$. So the difference with the constant removed is $36787.7601766572$ - already huge and still climbing - whereas the difference with $x/e$ removed is $0.1839389542$, within $7.7\times 10^{-7}$ of $\tfrac{1}{2e}$, matching the predicted error term $\tfrac{5}{24e\,x}=7.66\times 10^{-7}$.