## Infinite limits with cot?

1. The problem statement, all variables and given/known data
lim x->pi- cot(x)

2. Relevant equations
cot(x) = cos(x)/sin(x)

3. The attempt at a solution

so substituting pi into:
cot(pi) = cos(pi)/sin(pi)
= -1/0
so you have a negative over 0, approaching from the -ve side of pi wouldn't it be +infinity? why is it -infinity?

additionally this confuses me because a previous question I was working went like:

1)lim x->-3+ (x+2)/(x+3) = - infinity
2)lim x->-3- (x+2)/(x+3) = + infinity
when substituting in 3, one would get a -ve int/0.
so i thought you found out whether it is +ve or -ve infinity by multiplying signs.
1) -3+ so take + times - (from -ve int) = -ve ...and you get -ve infinity
2) -3- so take - times - (from -ve int) = +ve ...and you get +ve infinity

but that was the way a friend showed me, its worked for all the questions up until the cotx one. any help in understanding is much appreciated, thanks.

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Mentor
 Quote by shocklightnin 1. The problem statement, all variables and given/known data lim x->pi- cot(x) 2. Relevant equations cot(x) = cos(x)/sin(x) 3. The attempt at a solution so substituting pi into: cot(pi) = cos(pi)/sin(pi) = -1/0 so you have a negative over 0, approaching from the -ve side of pi wouldn't it be +infinity? why is it -infinity? additionally this confuses me because a previous question I was working went like: 1)lim x->-3+ (x+2)/(x+3) = - infinity 2)lim x->-3- (x+2)/(x+3) = + infinity when substituting in 3, one would get a -ve int/0. so i thought you found out whether it is +ve or -ve infinity by multiplying signs. 1) -3+ so take + times - (from -ve int) = -ve ...and you get -ve infinity 2) -3- so take - times - (from -ve int) = +ve ...and you get +ve infinity but that was the way a friend showed me, its worked for all the questions up until the cotx one. any help in understanding is much appreciated, thanks.
What is the sign of sin(x) when x is a little less than π ?

 positive..?

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## Infinite limits with cot?

Right, so it should be $$\frac{-1}{0^+}$$ because we're looking at $\sin(\pi ^-)$

 ooh right right! so thats why its -ve infinity. ah thanks, got it now :P

 Tags infinite limits, trig