# Thread: I did a find the limit of something involving trig and I'm not sure I am right.

1. ## I did a find the limit of something involving trig and I'm not sure I am right.

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.

2. Originally Posted by tastybrownies
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.
Clearly if @ --> 0 then -sin@/(cos@ + 1) --> -0/(1 + 1) = -0/2 = 0.