<p>i assume ur using IVT to prove the existence of roots, values where y is 0. so there are 3 things u must confirm with each one</p>
<ol>
<li><p>f is continuous on [a,b]. This is always true for a polynomial. For trig functions and rational functions, a f function is continuous on all parts in its domain. Just confirm that the interval given is a subset of the domain for 2 and 3.</p></li>
<li><p>f(a) < 0 < f(b). (f(a) and f(b) are of opposite signs. Just plug in your two endpoint values to confirm this.</p></li>
<li><p>f(a) does not equal f(b). If 2 is true, this is obviously true, but is stated for the sake of following the theorem.</p></li>
</ol>
<p>then ther ewill exist a root. if your not using this to prove roots, which you may be doing, im not sure, then just prove that the function is continuous and f(a) does not equal f(b). Then there exsits a number c in (a,b) such that f(c) = N <--- a solution. The IVT to prove roots assumes there exists a number c such that f(c) = N = 0. IVT is very useful ^_^</p>
<p>as far as limits. this is how i would do it. x csc5x = x / sin5x. then then multiply denom and num by 5 to get 5x / 5 sin5x. then, by some property 5x/sin5x wll cancel out so ull have 1/5 which is ur answer. </p>
<p>the law is sint/t = 0 or 1, which will cancel out. there is a law for cosines which escapes me atm...</p>
<p>Ummm, I can't help you with the above limit question, as I don't know what x approaches to, but I will assume x approaches 0</p>
<p>This seems to assume you know Calc already, is this Calc BC?
Plug in 0 for x and you get 0/0, this meets the conditions for L'Hopitals' Rule, which says if you get something of the form 0/0, infinity/infinity, and such, you can differentiate the top and differentiate the bottom (no quotient rule). until you get a defined limit. Differentiate 1 - sec^2 (2x) and you get (Oh, boy, what a lot of chain rule) -4sec^2(2x)tan(2x) and differentiate the bottom, and you get 8x. Now you have 4sec^2(2x)tan(2x)/(8x) and then you can plug in 0, which gets you 0/0 again!
There is no limit(lol) to how many times you can do L'Hopitals as long as you can get those forms everytime. So you do it again, and after a lot of pain and suffering differentiating the top part, you should get -8sec^4(2x)+ (something else but who cares, because there's a tan(2x) in there somewhere which evaluates to 0 when you plug in 0 for x, just a handy tip to make life easier:)), so basically we have just -8sec^4(2x) for the top part, since we omitted the unnecessary part, and 8 for the bottom part. Now, we have -8sec^4(2x)/8. Plug in 0 for x and we get -1 as the answer, which is our limit.
If you graph the function in question, you should see it gets closer to -1 as you go to 0.</p>
<p>Also, you can use L'Hopitals to solve the question about x csc5x. Since x csc5x = x/(sin(5x)), plug in 0 and you will see that you get 0/0, which means L'Hopitals applies. Differentiate the top, and you get 1. Differentiate the bottom, and you get 5cos(5x), so by L'Hopitals, the limit as x approaches 0 for x csc5x should be the same for 1/(5cos(5x)). So, plug in 0 for x and you get 1/5, which is the answer posted earlier.</p>
<p>im not that far in calculus to know L'Hopitals rule or w/e. but for that one u can do the cheap way of evaluating a limit. in your calculator try going to y= and plugging in (1- ((1/tan2x)^2) / (4x^2). assuming ur limit is as x approaches 0, select numbers very close to 0 and approaching 0. ur x values should be. -0.1, -0.01, -0.001, -0.0001, etc. and positive of those numbers. See wat they approach. However, I'm sure ur teacher wants u to do it the "proper" way, which i dont know how to do</p>
<p>"Also, you can use L'Hopitals to solve the question about x csc5x. Since x csc5x = x/(sin(5x)), plug in 0 and you will see that you get 0/0, which means L'Hopitals applies. Differentiate the top, and you get 1. Differentiate the bottom, and you get 5cos(5x), so by L'Hopitals, the limit as x approaches 0 for x csc5x should be the same for 1/(5cos(5x)). So, plug in 0 for x and you get 1/5, which is the answer posted earlier."</p>
<p>-that was the way I would have suggested but he seemed to be going INTO calculus...</p>
<p>also, aa6590, I don't think that would be a valid way to evaluate the limit on a test or something (at least in my class it wasn't)</p>