How would one approach finding this limit without using Taylor's series? $$\lim_{x \to 0} \frac{\tan{(\sin{(x)}})}{\sin{(\tan{(x)}})}$$
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1Just using $\sin x \approx x \approx \tan x$ is enough. That is the first term of a Taylor series for each. – Ross Millikan Nov 12 '18 at 22:06
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@RossMillikan Without Taylor's is requested. Of course, I agree, it can be equivalently derived from the standard limit. – user Nov 12 '18 at 22:09
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@gimusi: I admitted that, but $\sin x \approx x$ is a standard limit and we don't always call that the first term of the Taylor series (though it is). – Ross Millikan Nov 12 '18 at 22:17
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@RossMillikan Yes of course, the fact is that you were referring indeed to Taylor in your first comment. I thought you didn't see that limitation for that. – user Nov 12 '18 at 22:19
1 Answers
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HINT
By standard limits use that
$$\frac{\tan{(\sin{x}})}{\sin{(\tan{x}})} =\frac{\tan{(\sin{x}})}{\sin x} \frac{\tan x}{\sin{(\tan{x}})} \frac{\sin x}{x}\frac{x}{\tan x}$$
user
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@gimusi Luckily for me, that equation has already been proved in my class and I am able to freely use it. Thanks! – blahblah Nov 12 '18 at 21:57
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2@gimusi $(\frac{\sin\tan x^3}{\tan x^3}\frac{\tan x^3}{x^3}(\frac{x}{\sin x})^3)^2(\frac{\sin^2 x}{\tan\sin^2 x})^3$ – J.G. Nov 12 '18 at 22:18
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@Tomasz And what about $$\frac{\sin^3(\tan^2(\sin (\sqrt x)))}{\tan(\sin^2(\tan^{3/2} x))}$$ – user Nov 12 '18 at 22:28
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