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I am having trouble finding the limit for the following function:

$\lim_{x \to 0} f(x)=\frac{\tan 3x}{\tan 2x}$

I do not want to use L'Hospital's rule.

I know that $\frac{\tan 3x}{\tan 2x} = \frac{\sin 3x \cos 2x}{\sin 2x \cos 3x}$, but that does not help me.

Can you maybe give me a hint? Thanks.

Martin Sleziak
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Frithjof
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  • Do you know $\lim_{x\to 0} \frac{\tan x}x$? In that case, rewrite $f$ as $$ f(x) = \frac 23 \cdot \frac{\frac{3x}x}{\frac{2x}x} $$ – martini Oct 08 '15 at 14:53
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    See this almost identical problem: http://math.stackexchange.com/questions/1467848/limit-lim-x-to-0-frac-tan3x-sin2x/1467852#1467852 – Simon S Oct 08 '15 at 14:53

2 Answers2

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$$\frac{\sin 3x\cos 2x}{\sin 2x\cos 3x}=\frac{3}{2}.\frac{\sin 3x}{3x}.\frac{2x}{\sin 2x}.\frac{\cos 2x}{\cos 3x}$$

Empty
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Using $\tan (A+B)=\frac {\tan A+\tan B}{1-\tan A\tan B}$ twice we obtain $$\tan 2x =\frac {2\tan x}{1-\tan^2 x}; \tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$$ so that $$\frac {\tan 2x}{\tan 3x}=\frac {2(1-3\tan^2 x)}{(3-\tan^2 x)(1-\tan^2 x)}$$

And (having cancelled a factor of $\tan x$) nothing vanishes and the limit is easy to find, even if you know no other limits.

Mark Bennet
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