# Using different limit laws give different answer?

1. Sep 2, 2015

### FaroukYasser

Hi,

My instructor gave us a challenge problem to solve in limits: $\lim _{ x\rightarrow \infty }{ \left[ { x }^{ 2 }\left( 1-\cos { \frac { 5.1 }{ x } } \right) \right] }$ Note that we did not take Hospital's rule yet so we couldn't have used it.

Now my first thought was to use the product law of limits to get:

$\lim _{ x\rightarrow \infty }{ \left[ { x }^{ 2 }\left( 1-\cos { \frac { 5.1 }{ x } } \right) \right] } =\lim _{ x\rightarrow \infty }{ \left[ { x }^{ 2 } \right] } \cdot \lim _{ x\rightarrow \infty }{ \left[ \left( 1-\cos { \frac { 5.1 }{ x } } \right) \right] } \\ \lim _{ x\rightarrow \infty }{ \left[ { x }^{ 2 }\left( 1-\cos { \frac { 5.1 }{ x } } \right) \right] } =\lim _{ x\rightarrow \infty }{ \left[ { x }^{ 2 } \right] } \cdot 0\\ \lim _{ x\rightarrow \infty }{ \left[ { x }^{ 2 }\left( 1-\cos { \frac { 5.1 }{ x } } \right) \right] } =0$

But when I pressed submit, it gave me an incorrect answer. I tried playing with it algebraically and I ended up giving up and having to use L'Hospital rule, even though we haven't taken it yet. I did:

$\lim _{ x\rightarrow \infty }{ \left[ { x }^{ 2 }\left( 1-\cos { \frac { 5.1 }{ x } } \right) \right] } =\lim _{ x\rightarrow \infty }{ \left[ \frac { 1-\cos { \frac { 5.1 }{ x } } }{ \frac { 1 }{ { x }^{ 2 } } } \right] } \\ Using\quad Hospital's\quad rule:\\ \lim _{ x\rightarrow \infty }{ \left[ \frac { 1-\cos { \frac { 5.1 }{ x } } }{ \frac { 1 }{ { x }^{ 2 } } } \right] } =\lim _{ x\rightarrow \infty }{ \left[ 2.55x\sin { \frac { 5.1 }{ x } } \right] } \\ using\quad Hospital's\quad rule\quad again:\\ \lim _{ x\rightarrow \infty }{ \left[ 2.55x\sin { \frac { 5.1 }{ x } } \right] } =\lim _{ x\rightarrow \infty }{ \left[ 2.55\cdot 5.1\cdot \cos { \frac { 5.1 }{ x } } \right] } \\ =\lim _{ x\rightarrow \infty }{ \left[ \frac { 2601 }{ 200 } \sin { \frac { 5.1 }{ x } } \right] } =\frac { 2601 }{ 200 }$

Can anyone tell me why the first method didn't give me the correct answer? Also any hints about how to solve this algebraically or using any way other than L'Hospotal's rule?

Thanks

2. Sep 2, 2015

### Staff: Mentor

Isn't a basic premise of the product limit rule that all factors have limits?

In your case, the x^2 limit does not and so it can't be applied.

http://dl.uncw.edu/digilib/mathematics/calculus/limits/freeze/limitlaws.html

Can you use the half-angle identities ie sin^2(x) = 1/2 * (1 - cos(2x)) to simplify the limit?

http://www.purplemath.com/modules/idents.htm

This is a pretty interesting function for 0<x<1 it has sine like properties that goes to 0 at x=0 and for large x it looks like (13-1/x). You have to zoom out on the x-axis to see it.

Last edited: Sep 2, 2015
3. Sep 2, 2015

### FaroukYasser

I see now where the first method went wrong. However, I tried using the double angle formula and reached:

$\lim _{ x\rightarrow \infty }{ \left[ x^{ 2 }\left( 1-cos\left( \frac { 5.1 }{ x } \right) \right) \right] } =\lim _{ x\rightarrow \infty }{ \left[ 2x^{ 2 }\sin ^{ 2 }{ \left( \frac { 2.55 }{ x } \right) } \right] } \\ =2\lim _{ x\rightarrow \infty }{ \left[ x\sin { \left( \frac { 2.55 }{ x } \right) } \right] } ^{ 2 }=??$

But I seem to be stuck again. I think my problem is dealing with the x^2 so that I get rid of it without having a zero in the denominator and numerator (So I don't have to use L'Hospital rule)

Again, thanks a lot for your help.

4. Sep 2, 2015

### Qwertywerty

What is the value of sin(t)/t, if t→0 ?

Hope this helps.

5. Sep 2, 2015

### FaroukYasser

Hi, I know that the limit of sint/t as t approaches 0 is 1. I know a geometric proof to this as well as L'Hospital's method ie(lim(sint/t) = lim(cost/1). I am not sure how this could help me progress further taking into consideration that I don't want to use L'Hospital method if possible.

Sorry if my English isn't that good.

6. Sep 2, 2015

### Staff: Mentor

The other technique you need to use is a variable replacement say 2u= 5.1/x ie x = 5.1/(2 u) along with using the HALF-Angle identity.

This means you'll have to adjust the limit to be as u goes to zero.

Lastly, for very small u what is sin(u) approximately equal to (aka the small angle approximation)?

7. Sep 2, 2015

### FaroukYasser

Thanks a lot, finally figured it out :D

$2\lim _{ x\rightarrow \infty }{ \left[ x\sin { \left( \frac { 2.55 }{ x } \right) } \right] } ^{ 2 }\\ u=\frac { 2.55 }{ x } \quad \Rightarrow \quad x=\frac { 2.55 }{ u } \\ also\quad x\rightarrow \infty \quad when\quad u\rightarrow 0\\ \therefore \quad Limit\quad becomes:\\ 2\lim _{ u\rightarrow 0 }{ \left[ \frac { 2.55 }{ u } \sin { \left( u \right) } \right] } ^{ 2 }=2\cdot 2.55^{ 2 }\lim _{ u\rightarrow 0 }{ \left[ \frac { \sin { \left( u \right) } }{ u } \right] } ^{ 2 }\\ =\quad 2\cdot 2.55^{ 2 }\cdot 1\quad =\quad \frac { 2601 }{ 200 }$

The substitution idea was totally out of my head. Thanks again :)

8. Sep 2, 2015

### Staff: Mentor

There's a way that's quite a bit simpler: multiply (1 - cos(5.1/x)) by 1 in the form of its conjugate over itself, then make a substitution u = 5.1/x, and use the fact that $\lim_{u \to 0}\frac{\sin(u)}{u} = 1$. Note that if x → ∞, 1/x → ∞.

In a bit more detail,
$x^2(1 - \cos(5.1/x)) = x^2(1 - \cos(5.1/x)) \frac{1 + \cos(5.1/x)}{1 + \cos(5.1/x)} = x^2 \frac{1 - \cos^2(5.1/x)}{1 + \cos(5.1/x)} = x^2 \frac{\sin^2(5.1/x)}{1 + \cos(5.1/x)}$
The last expression can be split up into
$\frac{1}{1 + \cos(5.1/x)} \cdot x^2 \sin^2(5.1/x)$
The first factor has a limit as x → ∞, and the second factor can be manipulated via the substitution mentioned to a $\sin(u)/u$ limit. The resulting limit is as you show, or 13.005.

9. Sep 2, 2015

### FaroukYasser

Thanks for taking the time to answer :) I appreciate all the different solutions.