Proving that tan(x) is unbounded on [0, pi/2)

[torquerotates]

Homework Statement

Prove tan(x) is unbounded on [0,pi/2)

Homework Equations

if s is unbounded, s>n

The Attempt at a Solution

So I'm doing this by constructing a sequence. I don't know if this is valid. If tan(x) is unbounded on [0,pi/2), there exists x(n) in [0, pi/2) such that |tan(x(n))|>n for all n.

I can construct an sequence on [0,pi/2) such that pi/4<x1<x2<x3<...<x(n). such that

for n=1, tan(x1)>1

sin(x1)/cos(x1)>1 true since pi/4<x1<=pi/2

assume tan(x(n))>n is true.

well tan(x(n+1))=tan(e+x(n)) { since x(n+1)>x(n), I can say, x(n+1)=x(n)+e
for some number e}


=[tan(e)+ tan(x(n))]/[1-tan(e)tan(x(n)]<[tan(e)+n]/[1-tan(e)tan(x(n)]

taking lim[tan(e)+n]/[1-tan(e)tan(x(n)]=(0+n)/[1-0]=n
e->0

hence tan(x(n+1))>n

hence tan(x) is unbounded on [0,pi/2)

I don't even know if this is correct. I don't know if I can construct a sequence that does this and I don't know if I can take limits that way.

 

[hunt_mat]

Look at $$f(x)=\tan x-x$$ and show that it increasing but as a zero, this will show that after the zero $$\tan x>x$$ for that zero.

 

[Dickfore]

You need to formilze the fact that $\cos{\pi/2} = 0$, so $\tan{(x)}$ has a vertical asymptote there.

 

[hunt_mat]

Dickfore has a very good point, you could try and formalise an asymptotic series around pi/2.

 

[deluks917]

The easiest way is to show the bottom goes to zero as x goes to pi/2 while the top does not. Show that on some interval of (0,pi/2) sin(x) >= 1/2 (or some other non-zero number).