Find $\displaystyle\lim_{x \rightarrow 0} \frac{\tan x+\tan 2 x+\ldots+\tan n x}{\operatorname{arctg} x}$ [closed]
Solution 1:
By the asymptotic relations: $$\lim_{x\to 0}\frac{\tan(\alpha\cdot x)}{\alpha \cdot x}=1$$ And: $$\lim_{x\to 0}\frac{\arctan(x)}{x}=1$$
So, the numerator becomes: $$\tan(x)+\tan(2x)+\dots+\tan(nx)\sim x+2x+\dots+nx\,\,\, x\to 0$$
Thus the limit: $$\lim_{x \to 0} \frac{\tan x+\tan(2x)+\ldots+\tan(nx)}{\arctan(x) }=\lim_{x\to 0}\frac{x\cdot \frac{n\cdot (n+1)}{2}}{x}=\frac{n\cdot (n+1)}{2}$$