- #1
bluevires
- 20
- 0
Hey guys
I'm having trouble proving this limit using delta-epsilon definition
I would appreciate some help if possible
[tex]
\lim_{x\rightarrow \infty} \arctan(x)=\frac{\pi}{2}
[/tex]
I know that in order for the statement to be true,
Assuming
Epislon>0
Then |f(x)-L|<Epsilon for x> N
but i havn't had much experience working with trignometric functions, so I don't know how should I set my N equals to, and how could I convert that to |f(x)-L|< Epsilon
I'm having trouble proving this limit using delta-epsilon definition
I would appreciate some help if possible
[tex]
\lim_{x\rightarrow \infty} \arctan(x)=\frac{\pi}{2}
[/tex]
I know that in order for the statement to be true,
Assuming
Epislon>0
Then |f(x)-L|<Epsilon for x> N
but i havn't had much experience working with trignometric functions, so I don't know how should I set my N equals to, and how could I convert that to |f(x)-L|< Epsilon
Last edited: