# Limit involving infinity

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

limx→∞(1+1x)x2\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x^2}

It looks eerily similar to the limit case yielding ee 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}}\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}}\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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Decompose x^2=x\cdot xx^2=x\cdot x and try to obtain the limit for ee
– zar
Oct 20 at 22:57

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5

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For any k>1k>1, it is easy to se that for x>kx>k,

\left( 1+\frac1{x} \right)^{x^2} > \left( 1+\frac1{x} \right)^{kx} =\left( \left( 1+\frac1{x} \right)^{x} \right)^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
\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 >1k >1,
\lim_{x\to\infty}\left( 1+\frac1{x} \right)^{x^2} \to \infty\lim_{x\to\infty}\left( 1+\frac1{x} \right)^{x^2} \to \infty

You should have been taking the limit:

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

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

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

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

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

Now you have a proper form for L’Hأ´pital’s rule: \frac{0}{0}\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}\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^\inftye^\infty which diverges.

If a>0a>0 and nn is a positive integer, then (1+a)^n \ge 1+na.(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.(1+1/n)^{n^2} \ge 1 + n^2/n = 1 + n. Thus (1+1/n)^{n^2} \to \infty.(1+1/n)^{n^2} \to \infty. It’s a short walk from here to obtain (1+1/x)^{x^2} \to \infty(1+1/x)^{x^2} \to \infty as x\to \inftyx\to \infty through real values.

Since f(t)=\frac{1}{t}f(t)=\frac{1}{t} is positive and decreasing on \mathbb{R}^+\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} 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}\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 +\inftyx\to +\infty is straightforward to compute by comparison.

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\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}}} =