now perhaps it is a little easier?
Been struggling with doing trig limits amd this problem is driving me insane. We are suppose to work the problem to sinx/x format. The problem is lim as t---0 tan^2(3t) / 2t. im sure its easy but not to me.
i started with doing lim t-0 sin^2 (3t)/ cos^2(3t)(2t) and im lost. am i even on thr right track
Dang this problem has driven me up a wall. The answer is 0 according to the book. If I take tan^2(3t)/2t. Shouldn't I get sin^2(3t)/(2t)(cos^2(3t)). So in reality we have (sin3t)^2/(2t)(cos3t)^2. I'm sorry for being retarded. I just am looking for steps so I can see where I'm going off course.
Hello, psilver1!
Your first step is a good one: .
We'd like to use the theorem: .
So we will "hammer" our problem into that form . . . legally.
We have: .
. . and we want: .
How do we get it? .Multiply top and bottom by
. .
The function becomes: .
We have: .
A quick intuitive way to evaluate this limit might be helpful. First, recognize that cos(x) at x = 0 is equal to 1. Therefore in this case, you can treat tan(x) as sin(x), since you are dividing by a quantity that approaches 1.
So, after separating the limit into simpler ones with which we are familiar, we are left with
Basically the numerator in the original limit "overpowers" the denominator in how fast it approaches zero, similar to how