Does the limit exist of Lim coth[log(base(1/2))(tan^-1(x))] x->0+

Question

Lim coth[log(base(1/2))(tan^-1(x))] x->0+

Answer 1

Our task is to analyze the limit, as x goes to 0 from the right, of coth(log(arctan(x))), where the logarithm has base 1/2. Let's work from the inside out. As x goes to 0 from the right, arctan(x) also goes to 0 from the right (it's continuous there, so this is simple), and so we have reduced our task to evaluating the limit, as x goes to 0 from the right, of coth(log(x)). To better understand this logarithm, we use the change of base formula, and rewrite our log(with base 1/2) as ln(x)/ln(1/2). Since ln(1/2) < 0, we have log(x) going to positive infinity instead of negative infinity as x goes to 0 from the right, and therefore our limit becomes the limit, as x goes to (positive) infinity, of coth(x). Now, we just have to recall the definition of coth(x) as: (e^x + e^(-x))/(e^x - e^(-x)). As x goes to infinity, the e^(-x) terms go to zero, and we are left with the limit, as x goes to infinity, of (e^x + 0)/(e^x - 0) = 1, which is equal to 1.