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How can I find $$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}}\;?$$

I know $\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{n} = \exp (x)$ but I don't know how can I put the definition in this particular limit.

I know then, that $\lim_{n\rightarrow\infty}\big(1+\frac{x}n\big)=1$, but I don't think this is right to consider.

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5 Answers 5

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$$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}} = \lim_{n\rightarrow\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{{\frac{x}{n}}{\sqrt{n}}}$$ From $$\lim_{n\rightarrow\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]=e \quad \text{and} \quad \lim_{n\rightarrow\infty}{{\frac{x}{n}}{\sqrt{n}}}=0,$$ **, we get $$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}} = e^0=1$$

EDIT
I add the note bellow as my calculation was considered insufficiently justified

**and because the terms are positive, and we don't have an indeterminate case $0^0$ or $1^{\infty}$ or $\infty ^0,\;$

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    $\begingroup$ One needs an argument for "if $a_n\to a$ and $b_n\to b$ ($b_n,b\geqslant0, a_n,a>0$) then $a_n^{b_n}\to a^b$". $\endgroup$
    – user587192
    Nov 20, 2018 at 15:12
  • $\begingroup$ Or, if you prefer, put $t={n \over x}.$ The mentioned limit becomes $\lim_{t \to \infty} (1+{1\over t})^{t}.$ $\endgroup$
    – user376343
    Nov 20, 2018 at 15:18
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    $\begingroup$ What do you use for the "we get..." step in the last line? $\endgroup$
    – user587192
    Nov 20, 2018 at 15:19
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    $\begingroup$ @user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $\lim_n a_n^{b_n} $ into $(\lim_n a_n)^{(\lim_n b_n)}$ but this isn't necessarily justified. $\endgroup$
    – Jam
    Nov 20, 2018 at 15:26
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    $\begingroup$ From $$\lim_{n\to\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]=e \quad \text{and} \quad \lim_{n\to\infty}{{\frac{x}{n}}{\sqrt{n}}}=0,$$ we get $$\lim_{n\to\infty}\left(1+\frac{x}n\right)^{\sqrt{n}} =\lim_{n\to\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{{\frac{x}{n}}{\sqrt{n}}} \color{red}{ =\left[\lim_{n\to\infty}\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{\lim_{n\to\infty}{\frac{x}{n}}{\sqrt{n}}} } \color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified. $\endgroup$
    – user587192
    Nov 20, 2018 at 15:44
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Let $y=(1+\frac{x}{n})^{\sqrt{n}}$. Note that $$ \ln y=\sqrt{n}\ln(1+\frac{x}{n})=\frac{\ln(1+\frac{x}{n})}{1/\sqrt{n}}. $$ As $n\to\infty$, this is a limit of the form $0/0$. Thus, we may apply L'Hopital's rule to obtain that $$ \lim_{n\to\infty}\ln y=\lim_{n\to\infty}\frac{-\frac{x}{n^{2}}/(1+\frac{x}{n})}{-\frac{1}{2}n^{-3/2}}=\lim_{n\to\infty}\frac{2x}{\sqrt{n}+\frac{x}{\sqrt{n}}}=0 $$ Thus, $$ \lim_{n\to\infty}y=\lim_{n\to\infty}e^{\ln(y)}=e^{\left(\lim_{n\to\infty}\ln(y)\right)}=e^{0}=1. $$

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Hint:

Compute the limit of the log: $\;\sqrt n\log\Bigl(1+ \dfrac xn\Bigr)$ and use equivalence: $$\log(1+u)\sim_0 u.$$

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  • $\begingroup$ I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $\log(1+u)$ with $u$. Since $\log(1+u)\sim u$, $\lim_{u\to\infty}\frac{\log(1+u)}{u}=1$ so $\lim_{n\to\infty}\frac{\log(1+x/n)/\left(\frac{x}{n}\right)}{\sqrt{n}/{\left(\frac{x}{n}\right)}}=\frac{\lim(\ldots)}{\lim_{n\to\infty}\sqrt{n}{\left(\frac{n}{x}\right)}}$. $\endgroup$
    – Jam
    Nov 20, 2018 at 15:32
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Hint for $x\geq1$:

  1. The expression and therefore the limit is relatively easily seen to be greater than or equal to $1$
  2. The expression is for any $\epsilon>0$ eventually smaller than $\left(1+\frac xn\right)^{\epsilon n}$
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By L'Hopital, $$ \lim_{y\to\infty}\sqrt{y}\log(1+\frac{x}{y}) =\lim_{t\to 0}\frac{\log(1+tx)}{\sqrt{t}} =\lim_{t\to 0}\dfrac{\frac{x}{1+tx}}{\frac{1}{2\sqrt{t}}}=0. $$ Now one can use the continuity of the exponential function: $$ \lim_{n\to\infty}\exp\left[\sqrt{n}\log(1+\frac{x}{n})\right] =\exp\left[\lim_{n\to\infty}\sqrt{n}\log(1+\frac{x}{n})\right]=1. $$

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