Some Nice Limits

[Dave]

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 )

[bloodhound]

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

[Dave]

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).

[bloodhound]

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

[alt_f13]

What is lim?

 

[edit]

 

And how did you get to the second step, dave, if you don't mind explaining it quickly.

[Dave]

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

[fourier jr]

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

[dryan]

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.

[alt_f13]

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?

[bloodhound]

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.

[bloodhound]

Nicked from my analysis lecture notes.

[Dave]

-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.

[Dave]

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

[fourier jr]

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.

[Dave]

Yeah, it's a nice theorem; very useful for finding difficult limits when you can't see an approach.

[stevem]

© 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]

[Dave]

Indeed - it is quite a lot of work though, when you can differentiate it in a couple seconds.

[stevem]

I wrote it out in detail but you should be able to do most of it in your head