1) lim t->infinity ( sin 3t / 3t)

2) lim x->3+ (arc tan (1/3-x))

assuming you mean sin( 3t) / 3t.

1) is 0, as sin(3t) just ossilates, and the 1/t goes to 0, so it is 0

2) for the lim on the graph coming from the left or the right doesn't matter, we can just plug in the 3.

atan(1/3-3)=-1.2120

the graph is below

http://s5.postimg.org/qx10azmtj/untitled.png

is there anyway to show the work on how to get the solution to the arc tang problem ?

Could always make a big table with values right before and after 3, plug it in and see what value you're approaching. Can't really plug 3 in to see the answer since it would make the function undefined.

This post was edited by Malacant on Feb 19 2015 06:26pm

I believe you meant to say

Limit as x aproaches 3 of tan^-1(1/(3-x))
Not tan^-1(1/3-x)

And Limit as x aproaches 3 of tan^-1(1/(3-x)) is undefined.
This is clear because limit of our function as x aproaches 3 from the left Is pi/2 and from the right is -pi/2 and since these values are not equal the limit does not exist.

He probably typed it right. The limit at 3 doesn't exist, but since he said 3+ they probably wanted the limit coming from the right.

Ah missed the plus, typing and reading math forums on phone isn't always the best

Ahhhh, yes... i see, he meant



2) lim x->3+ (arc tan (1/(3-x)))

in this case, it changes the answer, as this is the limit as x goes to +-infinity.



This here and the proof is much harder than just looking at the graph.

They probably want the answer as -pi/4 since the limit coming in from the right gets closer and closer to -.785. If they want this in degree, use your reference angel for pi/4 which is 45 degrees then make it negative since its in quadrant IV.