tan=sin(x)/cos(x) is a good start.
How about putting
lim x&rarrow;pi+
as
x&rarrow; as (π/2+δ) as δ-> 0+.
tan(x)
=sin(x)/cos(x)
=sin(π/2+δ)/cos(π/2+δ)
=(sin(π/2)cos(δ)+cos(π/2)sin(δ))/ (cos(π/2)cos(δ)-sin(π/2)sin(δ))
setting sin(π/2)=1 and cos(π/2)=0,
=(cos(δ)+0*sin(δ))/ (0*cos(δ)-sin(δ))
=cos(δ)/(-sin(δ)
and setting cos(δ)=1, sin(δ)=0 as δ->+0
=1/(-0)
=-∞
Find the limit x->pi/2^+ tan(x)
I know the answer is -∞, but I don't understand why. When I simplify to sin(x)/cos(x), I get 1/0^+.
2 answers
π/2 is 90º
on the unit circle ... tan = y/x
at 90º, x = 0
dividing by zero gives the infinite limit
depending on the direction of approach to π/2, the sign of the limit can be +/-
on the unit circle ... tan = y/x
at 90º, x = 0
dividing by zero gives the infinite limit
depending on the direction of approach to π/2, the sign of the limit can be +/-