Limits of Arctan and Exponential Tangent Functions

[Weave]

Homework Statement

a.\lim_x -->infty arctan(x^2-x^4) =
b.\lim _x --->pi/2 e^tan (x) =

Homework Equations

Limit laws

The Attempt at a Solution

a. I know I can use the limit laws to take the limit of Arctan@infinity=1/2, but then, I don't know if I should factor out the (x^2-x^4)

b. Don't even know.

 

[IMDerek]

a. x^2-x^4 approaches negative infinity as x approaches positive infinity.

b. Look at both sides of the limit.

 

[Weave]

Ok, I am pretty sure b. is 0 because of the rule lim x-->infin e^X=0. We can substitute and look at the graph and it is 0.

Still not sure about a., taking the limit of Arctan@infinity=1/2 and (x^2-x^4) approaches -infinity so what then?

 

[IMDerek]

What is arctan x as x approaches negative infinity? It is not 1/2.

For b, consider that if the right and left sides of the limit don't agree, then there is no limit.

 

[Weave]

What is arctan x as x approaches negative infinity? It is not 1/2.

I am looking as it approaches +infinity(sorry if I didn't specify) and I am positive it is 1/2. Basicaly [lim x->inf. arctan=1/2]*[lim x->inf. (x^2-X^4)=+infinity}
right?

 

[Gib Z]

Did you get the hint about b? Try tan 89.999 degrees in your calculator, then then 90.0001, what do you see?

 

[HallsofIvy]

I am looking as it approaches +infinity(sorry if I didn't specify) and I am positive it is 1/2. Basicaly [lim x->inf. arctan=1/2]*[lim x->inf. (x^2-X^4)=+infinity}
right?

You appear to be "positive" about several things that are not true!
tan(1/2)= 0.5463, not infinity! Perhaps you forgot a \pi?

You also say

because of the rule lim x-->infin e^X=0

There is no such rule. The limit, as x goes to infinity of e^x is definitely NOT 0! Surely you have seen a graph of y= e^x!