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Using L'Hôpital's rule to find $$\lim_{x\to \infty} \left(\frac {\tan\beta x - \beta \tan x} {\sin\beta x - \beta \sin x}\right)$$ Where $\beta$ is a nonzero constant different from $\pm 1$.

I find this question weird because the limit does not confine to one of the forms that you can use L'Hôpital's rule. It is not one of the following form $\infty \over \infty$, $0 \over 0$, $\infty -\infty$, $0\times \infty$, $1^\infty$, $\infty^0$. So we cannot directly use the rule. How do I justify this question?

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    $\begingroup$ Are you sure it's not a typo and the limit is for $x\to0$? $\endgroup$ – egreg Mar 2 '17 at 8:18
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This limit does not exist. The problem isn't that the numerator and denominator have $x\to\infty$ limits whose ratio is an indeterminate form; the problem is that neither the numerator nor the denominator nor their ratio has a limiting behaviour for large $x$. For example, if $\beta\in\mathbb{Z}\backslash\left\{ 0,\,\pm 1\right\}$ the function has period $2\pi$. (If $\beta=0$ or $\beta=\pm 1$, the function does not exist for any $x$ due to being an indeterminate form, unless in those cases you define it at all $x$ with L'Hôpital or something similar.) A similar argument applies to any $\beta\in\mathbb{Q}\backslash\left\{ 0,\,\pm 1\right\}$; continuity then extends it to $\beta\in\mathbb{R}\backslash\left\{ 0,\,\pm 1\right\}$.

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By L'Hospital,

$$\lim_{x\to0} \frac {\tan\beta x - \beta \tan x} {\sin\beta x - \beta \sin x}=\lim_{x\to0}\frac {\dfrac\beta{\cos^2\beta x} - \dfrac\beta{\cos^2 x}} {\beta\cos\beta x - \beta \cos x}=\lim_{x\to0}\frac {\cos^2x - \cos^2\beta x} {\cos\beta x - \cos x}=-2.$$


By Taylor, the linear terms ($\beta x$) cancel each other, and the cubic terms yield

$$\frac{\dfrac{\beta^3x^3}{3}-\dfrac{\beta x^3}{3}}{-\dfrac{\beta^3x^3}{6}+\dfrac{\beta x^3}{6}}=-2.$$

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