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I'm struggling with the following limit.

$$\lim_{n\to ∞} (e^n-2^n)^{1/n}$$

First off, $(e^n-2^n)^{1/n} \le (e^n)^{1/n}$.

Secondly, since $\lim_{n\to ∞} (1+1/n)^{n}=e$, then $(e^n-2^n)^{1/n} \ge (1+1/n)^{n}$.

In addition, since $\lim_{n\to ∞} (e^n)^{1/n}=e$, then $\lim_{n\to ∞} (e^n-2^n)^{1/n}=e.$

Is my solution correct?

bijonne
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  • You second part doesn't make sense to me (the conclusion $\ge (1+\frac1n)^n$) but the limit is correct. – AlexR Nov 07 '14 at 15:18
  • The limit is certainly $e$, but I do not understand the argument you give. – André Nicolas Nov 07 '14 at 15:19
  • @AndréNicolas I was just trying to squeeze the original sequence inbetween the two sequences with more obvious limits. – bijonne Nov 07 '14 at 15:58
  • The finding two sequences to squeeze between is a good idea. You found something "above." The "below" part was not carried out. You could have used the fact that $2^n\lt 0.8e^n$ for $n\ge 1$, making $e^n-2^n\gt $(0.2)e^n$. Now take the $n$-th root. – André Nicolas Nov 07 '14 at 16:52

2 Answers2

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One can't say what is behind the first 'then'.

In this kind of problem, the systematic approach is to factor out the dominant term of each sums (here is is $e^n$ inside the parents, because $e > 2$).

Heuristically, you get $$ (e^n + 2^n)^{1/n}\simeq (e^n )^{1/n} = e $$ and more rigorously: $$ (e^n + 2^n)^{1/n} = \left[ e^n \left(1 + \left(\frac 2e\right)\right)^n\right]^{1/n} = e \left[ 1 + \left(\frac 2e\right)^n\right]^{1/n} $$now as $$ 0<\frac 2e < \frac 12 $$the term $$1 + \left(\frac 2e\right)^n$$ is bounded, and so $$ \left[ 1 + \left(\frac 2e\right)^n\right]^{1/n} \to 1 $$

mookid
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$$\ \lim_{n\to\infty}(e^n-2^n)^{\frac{1}{n}}=\lim_{n\to\infty}\left[e^n\left(1-\left(\frac{2}{e}\right)^n\right)\right]^{\frac{1}{n}}=$$ $$=\lim_{n\to\infty}e\cdot\left[1-\left(\frac{2}{e}\right)^n\right]^{\frac{1}{n}}$$

now, since $\ e>2$, the second term tends to 1, so you get that the limit is $\ e$

Traklon
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Mosk
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