[tex]\displaystyle\lim_{x\to0^+}(2x+1)\cot x=\lim_{x\to0^+}\frac{2x+1}{\tan x}[/tex]
As [tex]x\to0[/tex], the numerator approaches 1 while [tex]\tan x\to0[/tex], but since [tex]x\to0[/tex] from above we have [tex]\tan x>0[/tex], which suggests the limit is [tex]+\infty[/tex].
