Dave Posted June 2, 2004 Share Posted June 2, 2004 Had my Analysis II paper today, and I thought I'd share a nice question with you all Find: a) [math]\lim_{x\to 0} \left( \frac{\sin(2x)\tan(5x)}{3x^2} \right)[/math] b) [math]\lim_{x\to 0} \left( \frac{\sec^{2}(x) - 1}{2x^2} \right)[/math] c) [math]\lim_{x\to 4} \left( \frac{\sqrt{x} - 2}{\sqrt{2x+1} - 3\sqrt{x-3}} \right)[/math] Have fun, I'll post the answers in a bit (in case you were wondering, I think I just about did okay on it ) Link to comment Share on other sites More sharing options...
bloodhound Posted June 2, 2004 Share Posted June 2, 2004 for the first two. just use L'Hopital's Rule twice so for a) lim = 10/3 b)lim = 1/2 still doing c) i was lucky these didnt come up in my analysis exam Link to comment Share on other sites More sharing options...
Dave Posted June 2, 2004 Author Share Posted June 2, 2004 Yup first two are right, although there's a slightly more elegant way of doing it. [math]\frac{\sin(2x)\tan(5x)}{3x^2} = \frac{\sin(2x)\sin(5x)}{3\cos(5x)x^2}[/math]. Now you can see that this is equal to: [math]\frac{\sin(2x)}{x}\cdot \frac{\sin(5x)}{x}\cdot \frac{1}{3\cos(5x)} = \frac{\sin(2x)}{2x}\cdot \frac{\sin(5x)}{5x}\cdot \frac{10}{3\cos(5x)}[/math]. So the first bit -> 1, second bit -> 1 and third bit -> 10/3 so the limit is 10/3. Can use a similar method for the second one as well (subst sec2(x) - 1 = tan(x) and do some other trickery). Link to comment Share on other sites More sharing options...
bloodhound Posted June 2, 2004 Share Posted June 2, 2004 oh yeah, forgot about that. i have been used to using algebra of limits in addition and subtraction. dont think i have ever used ALG in limit of products Link to comment Share on other sites More sharing options...
alt_f13 Posted June 3, 2004 Share Posted June 3, 2004 What is lim? [edit] And how did you get to the second step, dave, if you don't mind explaining it quickly. Link to comment Share on other sites More sharing options...
Dave Posted June 3, 2004 Author Share Posted June 3, 2004 Just by splitting the fraction up. Like 1/6 = 1/3*1/2. lim stands for "limit" - i.e. the limit of x^2 as x->7 is 49. It's a little more complex than that though Link to comment Share on other sites More sharing options...
fourier jr Posted June 4, 2004 Share Posted June 4, 2004 The 1st two are obvious by L'Hopital's rule (as someone said) & while I haven't really tried the 3rd one I think I'd just multiply the denominator by its conjugate. (multiply by sqrt(2x+1) + 3*sqrt(x-3) / sqrt(2x+1) + 3*sqrt(x-3) ) ie multiply by 1, but you've got to pick the right 1. hehe Link to comment Share on other sites More sharing options...
dryan Posted June 4, 2004 Share Posted June 4, 2004 Haha! Just l'Hospital the last one too! Fourier really messed me up for a while; he got me to try conjugates... If you just think about it, taking derivitives for l'Hospital will give you some negative 1/2 powers and will flip flop parts from top and bottom to settle the problem. I'd show you, but it just gets confusing without proper notation, and I'll trust that Dave will do that. I got -3/14. Link to comment Share on other sites More sharing options...
alt_f13 Posted June 4, 2004 Share Posted June 4, 2004 Just by splitting the fraction up. Like 1/6 = 1/3*1/2. lim stands for "limit" - i.e. the limit of x^2 as x->7 is 49. It's a little more complex than that though What's the complex part? What are limits used for? Link to comment Share on other sites More sharing options...
bloodhound Posted June 4, 2004 Share Posted June 4, 2004 I have to admit , the first thing i tried was to rationalise the denominator as well, although that got me nowhere, so i just gave up. I thought if I used L'Hopital's it would just give me weird indices. Limits are used in differentiation, integrals, improper integrals and all sorts of stuff. Link to comment Share on other sites More sharing options...
bloodhound Posted June 4, 2004 Share Posted June 4, 2004 Nicked from my analysis lecture notes. Link to comment Share on other sites More sharing options...
Dave Posted June 4, 2004 Author Share Posted June 4, 2004 -3/14 is the right answer. I used L'Hopital's rule, which should give you the correct answer, I'll post a proof later. Link to comment Share on other sites More sharing options...
Dave Posted June 4, 2004 Author Share Posted June 4, 2004 What are limits used for? Limits are used for lots of things as bloodhound said; they actually define differentiation at a point. For example, if you have a function f(x) then the differential of the function at a point c is defined as: [math]\lim_{x\to c} \frac{f(x) - f©}{x-c}[/math] There's lots of other things you can do with them as well Link to comment Share on other sites More sharing options...
fourier jr Posted June 5, 2004 Share Posted June 5, 2004 Haha!Just l'Hospital the last one too! Fourier really messed me up for a while; he got me to try conjugates. yeah I should have gotten pencil & paper. L'Hopital's rule is a good swiss-army knife of a theorem. Link to comment Share on other sites More sharing options...
Dave Posted June 6, 2004 Author Share Posted June 6, 2004 Yeah, it's a nice theorem; very useful for finding difficult limits when you can't see an approach. Link to comment Share on other sites More sharing options...
stevem Posted June 6, 2004 Share Posted June 6, 2004 © can be done by using the conjugate and without using L'Hopital [math]\frac{\sqrt{x}-2}{\sqrt{2x+1}-3\sqrt{x-3}}[/math] [math]=\frac{(\sqrt{x}-2)(\sqrt{2x+1}+3\sqrt{x-3})}{(\sqrt{2x+1}-3\sqrt{x-3})(\sqrt{2x+1}+3\sqrt{x-3})}[/math] [math]=\frac{(\sqrt{x}-2)(\sqrt{2x+1}+3\sqrt{x-3})}{2x+1-9(x-3)}[/math] [math]=\frac{\sqrt{x}-2}{x-4}.\frac{\sqrt{2x+1}+3\sqrt{x-3}}{-7}=\frac{1}{\sqrt{x}+2}.\frac{\sqrt{2x+1}+3\sqrt{x-3}}{-7}[/math] [math]\to \frac{1}{4}.\frac{6}{-7}=-\frac{3}{14} \text{ as } x \to 4[/math] Link to comment Share on other sites More sharing options...
Dave Posted June 6, 2004 Author Share Posted June 6, 2004 Indeed - it is quite a lot of work though, when you can differentiate it in a couple seconds. Link to comment Share on other sites More sharing options...
stevem Posted June 6, 2004 Share Posted June 6, 2004 I wrote it out in detail but you should be able to do most of it in your head Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now