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Alright, so I had a midterm today where I had to evaluate the following limit:

$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x^2}$$

It looks eerily similar to the limit case yielding $e$ but I couldn't find a way to utilize that, so I took this approach:

$$\lim_{x\to\infty}e^{\ln\left(\frac{x+1}{x}\right)^{x^2}}$$

Which simplified to this using log properties:

$$\lim_{x\to\infty}\frac{(x+1)^{x^2}}{x^{x^2}}$$

However, once I get here I get a bit stuck. Would using L'Hopital's via natural logarithmic differentiation of both the numerator and denominator be an advisable next step, or is this approach ultimately fruitless altogether?

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  • $\begingroup$ Decompose $x^2=x\cdot x$ and try to obtain the limit for $e$ $\endgroup$ – zar Oct 20 '16 at 22:57
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For any $k>1$, it is easy to se that for $x>k$, $$ \left( 1+\frac1{x} \right)^{x^2} > \left( 1+\frac1{x} \right)^{kx} =\left( \left( 1+\frac1{x} \right)^{x} \right)^k $$ so $$\lim_{x\to\infty}\left( 1+\frac1{x} \right)^{x^2} > \left(\lim_{x\to\infty} \left( 1+\frac1{x} \right)^{x} \right)^k = e^k $$ and since this applies for any value of $k >1$, $$\lim_{x\to\infty}\left( 1+\frac1{x} \right)^{x^2} \to \infty$$

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You should have been taking the limit:

$$ \lim_{x\rightarrow \infty} x^2\ln\left(\frac{x + 1}{x}\right) $$

Since that gives $\infty \cdot 0$, it's indeterminant, but you need to change it--the easiest way is to put the $x^2$ in the denominator:

$$ \frac{\ln\left(\frac{x + 1}{x}\right)}{\frac{1}{x^2}} $$

Now you have a proper form for L'Hôpital's rule: $\frac{0}{0}$:

\begin{align} \lim_{x\rightarrow \infty}\frac{\ln\left(\frac{x + 1}{x}\right)}{\frac{1} {x^2}} =&\ \lim_{x\rightarrow \infty} \frac{-\frac{\frac{1}{x^2}}{\frac{x+1}{x}}}{-\frac{2}{x^3}} \\ =&\ \lim_{x\rightarrow \infty}\frac{\frac{1}{x(x+1)}}{\frac{2}{x^3}} \\ =&\ \lim_{x\rightarrow \infty}\frac{x^2}{2(x+1)} \end{align}

Since the numerator's degree is higher than the denominator's, this limit diverges, thus the original limit goes to $e^\infty$ which diverges.

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Since $f(t)=\frac{1}{t}$ is positive and decreasing on $\mathbb{R}^+$, $$ x^2\log\left(1+\frac{1}{x}\right)= x^2\int_{x}^{x+1}\frac{dt}{t}\geq \frac{x^2}{x+1} $$ so $$\left(1+\frac{1}{x}\right)^{x^2}\geq \exp\left(\frac{x^2}{x+1}\right)\geq e^{x-1}$$ and the limit as $x\to +\infty$ is straightforward to compute by comparison.

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If $a>0$ and $n$ is a positive integer, then $(1+a)^n \ge 1+na.$ This is easy to prove by induction. Therefore $(1+1/n)^{n^2} \ge 1 + n^2/n = 1 + n.$ Thus $(1+1/n)^{n^2} \to \infty.$ It's a short walk from here to obtain $(1+1/x)^{x^2} \to \infty$ as $x\to \infty$ through real values.

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{x \to \infty}\pars{1 + {1 \over x}}^{x^{2}} & = \exp\pars{\lim_{x \to \infty}\bracks{x^{2}\ln\pars{1 + {1 \over x}}}} = \exp\pars{\lim_{x \to 0^{+}}\bracks{{\ln\pars{1 + x} \over x^{2}}}} \\[5mm] & = \exp\pars{\lim_{x \to 0^{+}}\bracks{{1/\pars{1 + x} \over 2x}}} = \bbx{+\infty}\qquad\pars{~L'H\hat{o}pital\ Rule~} \end{align}

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