# Limits without l'Hôpital

Find: $$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$ I have factorized it in this manner in an attempt to use the formulae.

I have tried to use that for $x$ tending to $0$, $\dfrac{\sin x}{x} = 1$ and that $\dfrac{e^x - 1}x$ is also $1$.

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Is this homework? What formulae have you tried to use? –  Pacciu Jun 11 '12 at 20:30
I'm unclear on what your fraction is. Is it: $$\frac{\quad\frac{\sin x}{x} - \cos x\quad}{\quad 2x\left(\frac{e^{2x} - 1}{2x} - 1\right)\quad},$$ or is it $$\frac{\quad\frac{\sin x}{x-\cos x}\quad}{\quad 2x\left(\frac{e^{2x}-1}{2x-1}\right)\quad},$$ or something else? –  Arturo Magidin Jun 11 '12 at 20:32
Thanks for the editing @Pacciu. I didnt know the syntax. Sorry, I hope it's clear now. No this is not homework. Self study type. Thanks for helping me! –  siddhantsharan Jun 11 '12 at 20:41
It seems not to be the kind of limit you can solve using "only" those simple formulae... I mean, it mostly seems you have to use Taylor's formulae to solve that limit. –  Pacciu Jun 11 '12 at 20:45
I thought there might be a way. Thanks. –  siddhantsharan Jun 11 '12 at 20:49
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You are given

$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$

I guess you know

$$\lim_{x\to 0}\dfrac{\sin x}{x}=1$$

$$\lim_{x\to 0} \dfrac{e^{2x} - 1}{2x}=1$$

The most healthy way of solving this is using

$$\frac{\sin x}{x} = 1-\frac {x^2}{6}+o(x^2)$$

$$\frac{e^x-1}{x}=1+\frac x 2 +o(x^2)$$

$$\cos x = 1-\frac {x^2}{2}+o(x^2)$$

This gives

$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$

\eqalign{ & \mathop {\lim }\limits_{x \to 0} \;\frac{{1 - \dfrac{{{x^2}}}{6} + o({x^2}) - 1 + \dfrac{{{x^2}}}{2} - o({x^2})}}{{2x\left( {1 + x + o({x^2}) - 1} \right)}} \cr & \mathop {\lim }\limits_{x \to 0} \;\frac{{\dfrac{{{x^2}}}{3} + o\left( {{x^2}} \right)}}{{2{x^2} + 2xo\left( {{x^2}} \right)}} \cr & \mathop {\lim }\limits_{x \to 0} \;\frac{{\dfrac{1}{3} + \dfrac{{o\left( {{x^2}} \right)}}{{{x^2}}}}}{{2 + 2\dfrac{{o\left( {{x^2}} \right)}}{{{x^2}}}}} = \dfrac{1}{6} \cr} Note that

\eqalign{ & \frac{{o\left( {{x^2}} \right)}}{{{x^2}}} \to 0 \cr & \frac{{2o\left( {{x^2}} \right)}}{x} \to 0 \cr}

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$$\dfrac{\sin(x)}{x} = 1 - \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$ $$\cos(x) = 1 - \dfrac{x^2}{2!} + \mathcal{O}(x^4)$$ $$\exp(2x) = 1 + 2x + \dfrac{(2x)^2}{2!} + \mathcal{O}(x^3)$$ Hence, $$\dfrac{\dfrac{\sin(x)}{x} - \cos(x)}{\exp(2x) - 1 -2x} = \dfrac{-\dfrac{x^2}{3!} + \dfrac{x^2}{2!} + \mathcal{O}(x^4)}{2x^2 + \mathcal{O}(x^3)} = \dfrac{\dfrac{x^2}{3} + \mathcal{O}(x^4)}{2x^2 + \mathcal{O}(x^3)} = \dfrac{\dfrac13 + \mathcal{O}(x^2)}{2 + \mathcal{O}(x)}$$ Hence, $$\lim_{x \rightarrow 0} \dfrac{\dfrac{\sin(x)}{x} - \cos(x)}{\exp(2x) - 1 -2x} = \dfrac16$$
HINT: This limit has a very beautiful connection to another limit, namely $\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^2} = 0$ that may be elementarily proved, see here. It's worth to discover the connection on your own.
The connection between $\tan x$ and $\exp(x)$ is not possible by remaining within limits of real analysis. I wonder how we make use of the limit $\lim_{x \to 0}(\tan x - x)/x^{2} = 0$ in the denominator. –  Paramanand Singh Feb 21 at 16:58