Clearly if @ --> 0 then -sin@/(cos@ + 1) --> -0/(1 + 1) = -0/2 = 0.Hello everyone, I just did an even problem in my book for practice but am not sure whether I am right or not.
The original problem was this: lim as THETA goes to zero, (cos@ -1)/(sin@)
Notice the @'s are supposed to be theta's.
I started off by taking the conjugate, on top I ended up with
cos^2@ -1/(sin@)(cos@ +1), At this point I recognized there was a Pythagorean identity on top, so rewrote it as,
-sin^2@(sin@)(cos@+1), I was not sure where to go from here, but canceled one of the sin@ by virtue of the denominator. SO
-sin@/(cos@+1), I took the greatest denominator and applied it to top and bottom, and got, (-sin@/@)/(cos@/@ +1) THIS is where I am not sure whether or not to apply the @ UNDER the 1 or not, because then if I plugged in 0 it would equal infinity,which is probably wrong. So, the top and the bottom part go to 1 because of the limit rules, so I ended up with -1/2 as my final answer. Again, I am not sure if this is right or not. I appreciated anyone who takes their time to go through this.
This is a great website which I'm sure I will continue to use in the future.
Happy Easter everyone!
Tried to use the symbols at the bottom but it messed up my equation, sorry.