# Lim -> inf, x(sin(1/x)) = 1, why?

1. Apr 8, 2005

### ktpr2

I would've thought

$$\lim_{x \rightarrow \infty} x sin(\frac{1}{x})} = 0$$

because

$$\lim_{x \rightarrow \infty} x = \infty$$ and $$\lim_{x \rightarrow \infty} sin(\frac{1}{x})} = sin ( \lim_{x \rightarrow \infty} \frac{1}{x} = sin( 0)= 0$$ and $$\infty * 0 = 0$$

I begin to wonder if they should go back to teaching infestimals because in cases

2. Apr 8, 2005

### Imo

I'll quickly point out...zero times infinity is not always zero. Take for instance
$$\lim_{x \rightarrow \infty} \frac{x}{x}} = 1$$
but before simplification it's
$$\infty * 0$$

3. Apr 8, 2005

### Galileo

You are only allowed to split the limit into several limits (products, sums etc) if the individual limits exist. Clearly if one of them doesn't exist (infinity), you are no longer allowed to do so. Thus

$$\lim_{x \to a}(f(x)+g(x))=\lim_{x \to a}f(x)+\lim_{x \to a}g(x)$$
and
$$\lim_{x \to a}f(x)g(x)=\lim_{x \to a}f(x)\cdot \lim_{x \to a}g(x)$$
hold only if the limits for f(x) and g(x) exist.

$$\lim_{x \to 0}\frac{\sin(x)}{x}$$

4. Apr 8, 2005

### ktpr2

Can i mathematically write? :

$$\lim_{x \to 0} (\frac{\sin(x)}{x}) =$$

$$(\frac{\sin(x \to 0)}{x \to 0}) =$$

given that $$sin(x \to 0)$$ itself approaches $$(x \to 0)$$ as $$(x \to 0)$$,

$$(\frac{x \to 0}{x \to 0}) = 1$$

here I am replacing x with the symbol $$x \to 0$$ and using it as a algebraic variable.

Last edited: Apr 8, 2005
5. Apr 8, 2005

### dextercioby

Alebragic variables are a special kind of variables & need to be manipulated with care.

Daniel.

6. Apr 8, 2005

### ktpr2

er so that's a yes but be careful how i manipulate them? treating x in this way makes some limits much easier to evaluate.

7. Apr 8, 2005

### dextercioby

We have a thread especially on this "sinc" limit...It's around here.

Daniel.

8. Apr 8, 2005

### ktpr2

what thread are you referring to? a search on sinc limit only pulls up this thread.

9. Apr 8, 2005

### dextercioby

10. Apr 8, 2005

### Gokul43201

Staff Emeritus
What happens if you substitute y in the place of 1/x in the original problem ?

11. Apr 8, 2005

### Data

No. Consider

$$\lim_{x\rightarrow 0} \frac{1-\cos x}{x}$$

12. Apr 8, 2005

### dextercioby

What Data meant was that $1-\cos x$ approaches zero as $x^{2}$ (hence "faster") than $x$...

Daniel.

13. Apr 8, 2005

### p53ud0 dr34m5

l'hopital's rule can be applied to the limit. all you have to do is manipulate the problem so that you get a $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

$$\lim_{x \to \infty}xsin(\frac{1}{x})=\lim_{x \to \infty} \frac{sin( \frac{1}{x})}{\frac{1}{x}}=\frac{0}{0}$$

now, you should be able to apply l'hopital's rule and find why it is 1.

14. Apr 8, 2005

### xanthym

ktpr2:
Are you familiar with l'Hospital's Rule?? This rule applies to all "0/0" cases mentioned here so far. In a certain sense, the rule involves the "speed" with which the numerator and denominator approach 0 because the Rule uses derivatives to resolve the indeterminate "0/0" forms. (It can also be applied similarly to "∞/∞" forms.) Thus:

$$1: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {\sin(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}} \color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \sin(x) \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right )} \color{black}} } \ \ = \ \ \frac { \cos \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 1 \right) }$$

$$2: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {1 \, - \, \cos(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \left \{ 1 \, - \, \cos(x) \right \} \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right ) } \ \ = \ \ \frac { \sin \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 0 \right) }$$

~~

Last edited: Apr 9, 2005
15. Apr 8, 2005

### dextercioby

Daniel.

16. Apr 8, 2005

### dextercioby

Shame you used such a pretty writing for something incorrect.

Daniel.

17. Apr 8, 2005

### xanthym

Minor surgery performed to repair the previous hasty logical leap:

$$1: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {\sin(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}} \color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \sin(x) \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right )} \color{black}} } \ \ = \ \ \frac { \cos \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 1 \right) }$$

$$2: \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac {1 \, - \, \cos(x)} {x} \ \ = \ \ \ \ \lim_{\color{blue} \mathbf{x \rightarrow 0}}\color{black} \, \frac{ \displaystyle \left (\frac {d} {dx} \left \{ 1 \, - \, \cos(x) \right \} \right )} {\displaystyle \left ( \frac {d} {dx} \left( x \right) \right ) } \ \ = \ \ \frac { \sin \color{blue} \mathbf{ \left( 0 \right) }} {\color{blue} \mathbf{ \left( 1 \right) } } \ \ = \ \ \color{red} \mathbf{ \left( 0 \right) }$$

~~

Last edited: Apr 9, 2005
18. Apr 8, 2005

### dextercioby

U can't use the derivative of "sine" & "cosine",unless u adopt another method for computing them,other than the definition...

Daniel.

19. Apr 8, 2005

### xanthym

Nobody is trying to "prove" anything here (Msgs #14 & #17).
The l'Hospital's Rule is being offered as an efficient method for determining limits of indeterminate forms.

~~

20. Apr 8, 2005

### dextercioby

How do you prove that

$$\frac{d}{dx}\sin x=\cos x$$

?

Daniel.