Question

How do you find the limit of `tan^-1(1/x)` as `x->0^+`?

Answer 1

`lim_(x->0^+) arctan(1/x) = pi/2`

Explanation:

`lim_(x->0^+) arctan(1/x)`

substitute: `y= 1/x`, with `lim_(x->0^+) y(x) = +oo`

So:

`lim_(x->0^+) arctan(1/x) = lim_(y->+oo) arctan(y) = pi/2`