A limit of indeterminate

Question

$$\lim _{x\to \infty }\frac{\left(e^x+x\right)^{n+1}-e^{\left(n+1\right)x}}{xe^{nx}}\:= \:\:?$$

I tried to get it to a simpler form like this:

$$\lim _{x\to \infty }\frac{\left(e^x+x\right)^{n+1}-e^{\left(n+1\right)x}}{xe^{nx}}\:=\:\lim _{x\to \infty }\frac{e^{\left(n+1\right)x}\left(1+\frac{x}{e^x}\right)^{n+1}-e^{\left(n+1\right)x}}{xe^{nx}}=\lim _{x\to \infty }\frac{e^x}{x}\left(\left(1+\frac{x}{e^x}\right)^{n+1}-1\right)$$

Then i noted $\frac{e^x}{x}$ with t which tends to infinity also. Then i applyed the formula for $a^n-b^n$ (I've considered $1$ as $1^\left(n+1\right)$) and i got 1, but the answer is $n+1$. What have i missed ?

Answer 1

Use binomial expansion on the numerator : $$\left(e^x+x\right)^{n+1}-e^{(n+1)x} = \sum_{k=0}^n \binom{k}{n+1} e^{kx}x^{n+1-k}$$ Now the dominant term here is the term of number $k=n$, so $$\frac{\left(e^x+x\right)^{n+1}-e^{(n+1)x}}{xe^{nx}} \sim \frac{(n+1)xe^{nx}}{xe^{nx}}=n+1$$ An alternative method, starting with your factorization, is to use $(1+u)^\alpha-1\sim \alpha u$, so : $$\frac{e^x}{x}\left(\left(1+\frac{x}{e^x}\right)^{n+1}-1\right) \sim \frac{e^x}{x}.(n+1)\frac{x}{e^x}$$ which gives the same result.

Answer 2

$$\begin{align}\lim_{x\to \infty}\frac{e^x}{x}((1+\frac{x}{e^x})^{n+1} - 1) &= \lim_{t\to \infty}t((1+\frac{1}{t})^{n+1}-1) \\&= \lim_{t\to \infty}t(1+\frac{n+1}{t}-1 + o(\frac{1}{t}))\\&=n+1\end{align}$$

Answer 3

First multiply numerator and denominator by e,

We get $\dfrac{e(1+\frac{x}{e^x})^{n+1}-1}{x}. $ = $$\dfrac{e*\left(e^{\frac{x}{e^x}(n+1)}-1\right)}{x}. $$ ( as $1^\infty$ form) Applying L' hopital's rule and simplifying,we get

= $$ \dfrac{e(e^{\frac{x}{e^x}(n+1)})*(n+1)*(1-x)}{e^x} = $$e(e^{\frac{x}{e^x}(n+1)-1})(n+1)$$

$$~e * e^{-1} * (n+1) = n+1$$