I'm trying to calculate $$\lim_{x\to0}\frac{\tan(5x)}{\tan(11x)}.$$ It seems simple but I cannot figure it out. Should $\tan$ be converted to $\sin$ and $\cos$?
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$\begingroup$ What is $\lim_{x \to 0} \frac{\tan ax}{ax}$? $\endgroup$– Тимофей ЛомоносовOct 17, 2013 at 6:12
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4 Answers
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$$\sin(x)\sim \tan(x)\sim x$$ when $x \to 0$.
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$$\lim_{y\to0}\frac{\tan y}y=\lim_{y\to0}\frac{\sin y}y\cdot\frac1{\lim_{y\to0}\cos y}$$
We know, $\displaystyle\lim_{y\to0}\frac{\sin y}y=1$
$$\text{Now,}\frac{\tan5x}{\tan11x}=\frac5{11}\cdot\frac{\frac{\sin5x}{5x}}{\frac{\sin11x}{11x}}\cdot\frac1{\cos5x}\cdot {\cos11x}$$
answered Oct 17, 2013 at 6:12
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You want to use the fact that $tanx = \frac{sinx}{cosx}$ and then apply the limit identitiy that $$ \lim_{x \to 0} \frac{sin\theta}{\theta} = 1$$ Here, you'd have to use $\theta = 5x$ and $\theta=11x$ when you slit up the two tangent terms. Hope this helps!
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Use L'Hopitals rule since when you substitute $0$ into $\dfrac{\tan(5x)}{\tan(11x)}$ gives you the indeterminate form of $\dfrac{0}{0}$ Thus L'Hopital rule will work.
