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

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 ?

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3 Answers 3

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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.

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  • $\begingroup$ The first tip worked well, but i still wonder why it doesn't work to apply $a^n - b^n$ on my last form. $\frac{1}{t}$ will tend to 0 and it will be $1*(1+n*0)$ $\endgroup$
    – Liviu
    Jan 15, 2017 at 16:16
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    $\begingroup$ It works if you don't forget $a=1+\frac{x}{e^x}$, so limit is $1$, not $0$ :-) $\endgroup$ Jan 15, 2017 at 18:03
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$$\begin{align}\lim_{x\to \infty}\frac{e^x}{x}((1+\frac{x}{e^x})^{n+1} - 1) &= \lim_{t\to \infty}t((1+\frac{1}{t})^{n+1}-1) \\&= \lim_{t\to \infty}t(1+\frac{n+1}{t}-1 + o(\frac{1}{t}))\\&=n+1\end{align}$$

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  • $\begingroup$ @SimpleArt I made and substitution so as $x \to \infty $ , $t \to \infty$ also. After that I use Taylor series for $(1+\frac{1}{t})^{n+1}$, because $\frac{1}{t} \to \infty $. $\endgroup$
    – openspace
    Jan 15, 2017 at 15:59
  • $\begingroup$ Why ? If sequence $a_{n}$ converges than it could be represented as $a + o(1)$, where a is limit. $\endgroup$
    – openspace
    Jan 15, 2017 at 16:03
  • $\begingroup$ Oh, my bad. Using the $+1$ and $-1$ simultaneously was hard to read. $\endgroup$ Jan 15, 2017 at 16:04
  • $\begingroup$ @SimpleArt my bad with English stylo :) $\endgroup$
    – openspace
    Jan 15, 2017 at 16:05
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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)$$

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

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