| TheLivingShadow said: Alright, I'm gonna assume it's the second case. Look: lim(x->0) (x+tan(x))/sin(x) = lim(x->0) (x/sin(x)) + lim(x->0) (tan(x)/sin(x)). What I did here was separate the fraction sum. Now, recall that tan(x)=sin(x)/cos(x). Then: = lim(x->0) (x/sin(x)) + lim(x->0) [(sin(x)/cos(x))/sin(x)]. The bold term is 1 (from the first limit identity you wrote) and the italics cancel out. Then the expression turns into: = 1 + lim(x->0)(1/cos(x)). But note that as x goes to 0, cos(x) goes to 1. So: = 1 + 1/1 = 2 Your answer is 2. |
okay my notes for that particular problem now make sense!
okay now I have:
lim x->0 3(1-cos(x))/x
well since I know 1-cos(x)/x=0 I seperate that from 3 to conclude 3 * 0=0
lim θ-> cos θ tan θ/ θ
Now for this one, θ is giving me problems because I don't know what to do with them.
lime x->0 tan^2 (x)/x
The square is really bothering me.
and thanks again for helping me, you don't know how much I appreciate it. This homework could really either help or destroy me.
