# Showing $\lim_{m\rightarrow\infty} \left(1-\frac1{m^2}\right)^m=1$

I want to prove $$\lim_{m\rightarrow\infty} \left(1-\frac1{m^2}\right)^m=1$$ without using the fact that $\lim_{m\rightarrow\infty}\left(1+\frac1m\right)^m=\mathrm e$.

I know by the Bernoulli-Inequality $$\left(1-\frac1{m^2}\right)^m\geq1-\frac1m$$ But now I don't know how to show $\left(1-\frac1{m^2}\right)^m\leq1$ for all $m\in\mathbb N$.

Anybody could help? Thanks.

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Hint: With $m$ an integer, use binomial theorem and then estimate the terms of if $(1-\frac{1}{m^2})^m$. –  Thomas Andrews Nov 20 '12 at 19:18
I don't see how $$\left(1-\frac1{m^2}\right)^m\geq1$$ is ever true for $m$ a positive real number. Try $m=2$ and it doesn't work. –  Thomas Andrews Nov 20 '12 at 19:20
@ThomasAndrews oh I've meant $1-\frac1m$ on the right side. –  user50120 Nov 20 '12 at 19:24

When you get stuck, try proving something harder instead. Perhaps

$$\left( 1-\frac1{a}\right)^n \le 1$$ for all $a\ge 1$ and $n\ge 1$?

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+1 Easy and straightforward. –  Babak S. Nov 20 '12 at 19:13
On the other hand, I'm not sure I understand the question, because $(1-\frac 1{m^2})^m\ge 1$ (which the OP "knows by the Bernoulli-Inequality") is definitely not true in the arithmetic I'm used to. For $m=1$ I get $0\ge 1$ and for $m=2$ I get $\frac{9}{16}\ge 1$. –  Henning Makholm Nov 20 '12 at 19:19
@HenningMakholm Yeah, he corrected to it $\geq 1-\frac{1}m$ –  Thomas Andrews Nov 20 '12 at 19:31

$$\left(1-\frac{1}{m^2}\right)^m=\left(1-\frac{1}{m}\right)^m\times \,\,\left(1+\frac{1}{m}\right)^m$$

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'reading the post' is your friend ;) although this one is the easiest solution... –  user50120 Nov 20 '12 at 19:38

Apply $\ln$, rewrite with $1/m$ in the denominator and then use L'Hospital's rule to bring the expression into the form of a rational function.

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$$\lim_{m \rightarrow \infty}(1-\frac{1}{m^2})^m=\lim_{m\rightarrow \infty}e^{\ln[1-\frac{1}{m^2}]^m}=\lim_{m\rightarrow \infty}e^{m\ln[1-\frac{1}{m^2}]},$$ the expression in the exponent tends to $\infty\cdot 0$, so we use L'Hopital's Rule:

$$\lim_{m\rightarrow \infty}e^{m\ln[1-\frac{1}{m^2}]}=\lim_{m \rightarrow \infty}e^{\frac{\ln[1-\frac{1}{m^2}]}{m}}=\lim_{m \rightarrow \infty}e^{\frac{2}{m^3-m}}=e^0=1$$

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