# Some Nice Limits

## Recommended Posts

Had my Analysis II paper today, and I thought I'd share a nice question with you all

Find:

a) $\lim_{x\to 0} \left( \frac{\sin(2x)\tan(5x)}{3x^2} \right)$

b) $\lim_{x\to 0} \left( \frac{\sec^{2}(x) - 1}{2x^2} \right)$

c) $\lim_{x\to 4} \left( \frac{\sqrt{x} - 2}{\sqrt{2x+1} - 3\sqrt{x-3}} \right)$

Have fun, I'll post the answers in a bit

(in case you were wondering, I think I just about did okay on it )

##### Share on other sites

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

##### Share on other sites

Yup first two are right, although there's a slightly more elegant way of doing it.

$\frac{\sin(2x)\tan(5x)}{3x^2} = \frac{\sin(2x)\sin(5x)}{3\cos(5x)x^2}$.

Now you can see that this is equal to:

$\frac{\sin(2x)}{x}\cdot \frac{\sin(5x)}{x}\cdot \frac{1}{3\cos(5x)} = \frac{\sin(2x)}{2x}\cdot \frac{\sin(5x)}{5x}\cdot \frac{10}{3\cos(5x)}$.

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).

##### Share on other sites

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

##### Share on other sites

What is lim?

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

##### Share on other sites

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

##### Share on other sites

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

##### Share on other sites

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.

##### Share on other sites
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?

##### Share on other sites

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.

##### Share on other sites

Nicked from my analysis lecture notes.

##### Share on other sites

I used L'Hopital's rule, which should give you the correct answer, I'll post a proof later.

##### Share on other sites
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:

$\lim_{x\to c} \frac{f(x) - f©}{x-c}$

There's lots of other things you can do with them as well

##### Share on other sites
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.

##### Share on other sites

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

##### Share on other sites

© can be done by using the conjugate and without using L'Hopital

$\frac{\sqrt{x}-2}{\sqrt{2x+1}-3\sqrt{x-3}}$

$=\frac{(\sqrt{x}-2)(\sqrt{2x+1}+3\sqrt{x-3})}{(\sqrt{2x+1}-3\sqrt{x-3})(\sqrt{2x+1}+3\sqrt{x-3})}$

$=\frac{(\sqrt{x}-2)(\sqrt{2x+1}+3\sqrt{x-3})}{2x+1-9(x-3)}$

$=\frac{\sqrt{x}-2}{x-4}.\frac{\sqrt{2x+1}+3\sqrt{x-3}}{-7}=\frac{1}{\sqrt{x}+2}.\frac{\sqrt{2x+1}+3\sqrt{x-3}}{-7}$

$\to \frac{1}{4}.\frac{6}{-7}=-\frac{3}{14} \text{ as } x \to 4$

##### Share on other sites

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

##### Share on other sites

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

## Create an account

Register a new account