$$\lim_{x\to0}\frac{\arctan x}{e^{2x}-1}$$ How to do this without L'Hôpital and such? $\arctan x=y$, then we rewrite it as $\lim_{y\to0}\frac y{e^{2\tan y}-1}$, but from here I'm stuck.
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4$$\lim_{x\to 0}\frac{1}{2}\cdot\frac{\arctan x}{x}\cdot\frac{2x}{e^{2x}-1}=\color{red}{\frac{1}{2}}$$ since $$\lim_{z\to 0}\frac{\tan z}{z}=\lim_{z\to 0}\frac{e^z-1}{z}=1.$$ – Jack D'Aurizio Oct 01 '16 at 14:42
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1@JackD'Aurizio And if the OP is wondering about these last two limits, reading them as two derivatives at $0$ may clear any confusion. – Clement C. Oct 01 '16 at 14:47
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1@ClementC. And for those without a background in derivatives, I've presented a "back-to-basics" way forward. ;-)) – Mark Viola Oct 01 '16 at 15:05
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I thought it might be instructive to present a way forward that goes back to "basics." Herein, we rely only on elementary inequalities and the squeeze theorem. To that end, we proceed with a primer.
PRIMER ON A SET OF ELEMENTARY INEQUALITIES:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$
for $x<1$.
And in THIS ANSWER, I showed using only elementary inequalities from geometry that the arctangent function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{|x|}{\sqrt{1+x^2}}\le |\arctan(x)|\le |x|} \tag 2$$
for all $x$.
Using $(1)$ and $(2)$ we can write for $1>x>0$
$$\frac{x}{\sqrt{1+x^2}\left(\frac{2x}{1-2x}\right)}\le \frac{\arctan(x)}{e^{2x}-1}\le \frac{x}{2x} \tag 3$$
whereupon applying the squeeze theorem to $(3)$, we find that
$$\lim_{x\to 0^+}\frac{\arctan(x)}{e^{2x}-1}=\frac12$$
Similarly, using $(1)$ and $(2)$ for $x<0$ we can write
$$\frac{x}{\left(\frac{2x}{1-2x}\right)}\le \frac{\arctan(x)}{e^{2x}-1}\le \frac{x}{\sqrt{1+x^2}\,\left(2x\right)} \tag 4$$
whereupon applying the squeeze theorem to $(4)$, we find that
$$\lim_{x\to 0^-}\frac{\arctan(x)}{e^{2x}-1}=\frac12$$
Inasmuch as the limits from the right and left sides are equal we can conclude that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\arctan(x)}{e^{2x}-1}=\frac12}$$
Mark Viola
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Another way using Taylor series $$\tan^{-1}(x)=x-\frac{x^3}{3}+O\left(x^4\right)$$ $$e^{2x}-1=2 x+2 x^2+\frac{4 x^3}{3}+O\left(x^4\right)$$ $$\frac{\tan^{-1}(x) } {e^{2x}-1 }=\frac{x-\frac{x^3}{3}+O\left(x^4\right) } {2 x+2 x^2+\frac{4 x^3}{3}+O\left(x^4\right) }=\frac{1-\frac{x^2}{3}+O\left(x^3\right) } {2 +2 x+\frac{4 x^2}{3}+O\left(x^3\right) }$$ Performing the long division $$\frac{\tan^{-1}(x) } {e^{2x}-1 }=\frac{1}{2}-\frac{x}{2}+O\left(x^3\right)$$ which shows the limit and how it is approached.
Claude Leibovici
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As a small comment, expansion up to $o (x) $ would have been enough here for the limit (although it's always simpler in hindsight). But it would not give the rate of decay of the additive term. – Clement C. Oct 01 '16 at 16:37
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@ClementC..I fully agree with you. I think that http://matheducators.stackexchange.com/questions/8339/teaching-limits-and-asymptotics-at-the-same-time could better explain my point of view. When I was teached limits (so long time ago !), almost immediately after they explain how to study the approach to the limit. Tis is why, when I use Taylor series, I use an extra term and explain that, for the same price, we have the limit and more. Cheers. – Claude Leibovici Oct 01 '16 at 18:04
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I agree with the comment -- simply meant to point out to readers that for the limit, one does not need all the terms. From an "experimental" point of view as well, it makes sense to get that extra term: at first, it's not always clear what order is sufficient, so better sewat one extra drop and have more than enough than doing too little and have to start from scratch after realizing there was just one pesky term missing in the development... – Clement C. Oct 01 '16 at 18:25