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

  • Thread starter ktpr2
  • Start date
It is well known that those functions are continuous, differentiable and have well known derivatives. No one has to "prove" that here. The OP was asking for an evaluation of the given limit. I provided a simple way to do that. I did not provide a course in analysis. He did not ask for one. If you wish to provide a more rigorous treatment, then feel free to do so. But, please don't criticize me for not doing so.In summary, the conversation discusses the use of l'Hospital's Rule to evaluate limits of indeterminate forms, specifically the limit of x*sin(1/x) as
  • #1
ktpr2
192
0
I would've thought

[tex] \lim_{x \rightarrow \infty} x sin(\frac{1}{x})} = 0 [/tex]

because

[tex] \lim_{x \rightarrow \infty} x = \infty [/tex] and [tex]\lim_{x \rightarrow \infty} sin(\frac{1}{x})} = sin ( \lim_{x \rightarrow \infty} \frac{1}{x} = sin( 0)= 0[/tex] and [tex] \infty * 0 = 0 [/tex]

I begin to wonder if they should go back to teaching infestimals because in cases
 
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  • #2
I'll quickly point out...zero times infinity is not always zero. Take for instance
[tex] \lim_{x \rightarrow \infty} \frac{x}{x}} = 1 [/tex]
but before simplification it's
[tex] \infty * 0 [/tex]
 
  • #3
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

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

To solve your limit, try to find a link with:

[tex]\lim_{x \to 0}\frac{\sin(x)}{x}[/tex]
 
  • #4
Can i mathematically write? :

[tex]\lim_{x \to 0} (\frac{\sin(x)}{x}) = [/tex]

[tex] (\frac{\sin(x \to 0)}{x \to 0}) = [/tex]

given that [tex] sin(x \to 0)[/tex] itself approaches [tex](x \to 0) [/tex] as [tex] (x \to 0)[/tex],

[tex] (\frac{x \to 0}{x \to 0}) = 1[/tex]

here I am replacing x with the symbol [tex] x \to 0[/tex] and using it as a algebraic variable.
 
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  • #5
Alebragic variables are a special kind of variables & need to be manipulated with care.
The answer is still 1,though...

Daniel.
 
  • #6
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
We have a thread especially on this "sinc" limit...It's around here.

Daniel.
 
  • #8
what thread are you referring to? a search on sinc limit only pulls up this thread.
 
  • #10
What happens if you substitute y in the place of 1/x in the original problem ?
 
  • #11
ktpr2 said:
Can i mathematically write?...

No. Consider

[tex]\lim_{x\rightarrow 0} \frac{1-\cos x}{x}[/tex]
 
  • #12
What Data meant was that [itex] 1-\cos x [/itex] approaches zero as [itex] x^{2} [/itex] (hence "faster") than [itex]x [/itex]...

Daniel.
 
  • #13
l'hopital's rule can be applied to the limit. all you have to do is manipulate the problem so that you get a [itex]\frac{0}{0}[/itex] or [itex]\frac{\infty}{\infty}[/itex].

[tex]\lim_{x \to \infty}xsin(\frac{1}{x})=\lim_{x \to \infty} \frac{sin( \frac{1}{x})}{\frac{1}{x}}=\frac{0}{0}[/tex]

now, you should be able to apply l'hopital's rule and find why it is 1.
 
  • #14
dextercioby said:
What Data meant was that [itex] 1-\cos x [/itex] approaches zero as [itex] x^{2} [/itex] (hence "faster") than [itex]x [/itex]...

Daniel.
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:


[tex] 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) } [/tex]


[tex] 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) } [/tex]



~~
 
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  • #15
It's logically incorrect to apply l'Hôpital's rule to those 2 limits...:wink: Read the thread i made a link to in post #9.In that thread u'll find another link to an illuminating wikipedia page.

Daniel.
 
  • #16
Shame you used such a pretty writing for something incorrect.

Daniel.
 
  • #17
Minor surgery performed to repair the previous hasty logical leap:


[tex] 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) } [/tex]


[tex] 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) } [/tex]


~~
 
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  • #18
U didn't read the thread,nor the wiki page,did u?

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

Daniel.
 
  • #19
dextercioby said:
U didn't read the thread,nor the wiki page,did u?

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

Daniel.
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
How do you prove that

[tex] \frac{d}{dx}\sin x=\cos x [/tex]

?

Daniel.
 
  • #21
dextercioby said:
How do you prove that

[tex] \frac{d}{dx}\sin x=\cos x [/tex]

?

Daniel.
The 2 equations in MSGs #14 & #17 show applications of l'Hospital's Rule for evaluating indeterminate limits. The purpose is merely to provide an efficient technique for doing so.
It is not the intention (in those msgs) to provide a rigorous approach, nor one that is based on fundamental derivations. Rather, the suggested use of l'Hospital's Rule is made only to allow for quick determination of indeterminate limit forms.





~~
 
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  • #22
I begin to wonder if they should go back to teaching infestimals because in cases

It would offer no help -- the nonstandard proof would also be done by applying the fact that [itex]\lim_{x \rightarrow 0} (\sin x)/x = 1[/itex].
 
  • #23
umm, you want a proof? let's see if i still remember the definition of a derivative

[tex]f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}[/tex]

so to prove:
[tex]\frac{d}{dx}sin(x)=cos(x)[/tex]

[tex]\lim_{h \to 0}\frac{sin(x+h)-sin(x)}{h}[/tex]

now, you need trig identities:
[tex]sin(a+b)=sin(a)cos(b)+sin(b)cos(b)[/tex]

[tex]\lim_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}[/tex]

[tex]\lim_{h \to 0}\frac{sin(x)(cos(h)-1)+sin(h)cos(x)}{h}[/tex]
[tex]\lim_{h \to 0}sin(x)*\lim_{h \to 0}\frac{cos(h)-1}{h}+\lim_{h \to 0}cos(x)*\lim_{h \to 0}\frac{sin(h)}{h}[/tex]

so:
[tex]sin(x)*0 + cos(x)*1 = cos(x)[/tex]
is that the proof you were looking for? i think its right; i don't really remember chapter two--chapter two was september. if so, then it is valid to use l'hopital's rule for his limit.
 
  • #24
EDIT- this is not directly realated to my original post, but it's very similar. I wished to find the limit as x -> 0 for sin (5x)/x.

i spoke to my teacher about this and he said to substitute u for 5x. What he thereafter did made sense except for one crucial part where he extracted 5 from u. I have attempted to explain below, w/o using a variable u, and would appreciate a look.

(btw, we have not covered l'Hospital's and it appears to be irrelevant in this case)

[tex]\lim_{x \to 0} (\frac{\sin(5x)}{x}) = [/tex]

we apply the limit to both parts of the fraction...

[tex] \frac{ sin( \lim_{x \to 0} 5x)}{ \lim_{x \to 0} x} = [/tex]

and use the rule that let's me "spit" the constant outside of the limit in question.

[tex] 5 ( \frac{ sin( \lim_{x \to 0} x)}{ \lim_{x \to 0} x} ) = [/tex]

(put the limits back outside)

[tex] 5 \lim_{x \to 0} ( \frac{ sin( x)}{ x} ) = [/tex]

and knowing that sin x/x as x --> 0 equals 1, I can finally write

[tex] 5 ( 1 ) = 5[/tex]

So, is this what's really going on?
 
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  • #25
ktpr2 said:
EDIT- this is not directly realated to my original post, but it's very similar. I wished to find the limit as x -> 0 for sin (5x)/x.

i spoke to my teacher about this and he said to substitute u for 5x. What he thereafter did made sense except for one crucial part where he extracted 5 from u. I have attempted to explain below, w/o using a variable u, and would appreciate a look.

(btw, we have not covered l'Hospital's and it appears to be irrelevant in this case)

[tex]\lim_{x \to 0} (\frac{\sin(5x)}{x}) = [/tex]

we apply the limit to both parts of the fraction...

[tex] \frac{ sin( \lim_{x \to 0} 5x)}{ \lim_{x \to 0} x} = [/tex]

and use the rule that let's me "spit" the constant outside of the limit in question.

[tex] 5 ( \frac{ sin( \lim_{x \to 0} x)}{ \lim_{x \to 0} x} ) = [/tex]

(put the limits back outside)

[tex] 5 \lim_{x \to 0} ( \frac{ sin( x)}{ x} ) = [/tex]

and knowing that sin x/x as x --> 0 equals 1, I can finally write

[tex] 5 ( 1 ) = 5[/tex]

So, is this what's really going on?
Since {sin(5*θ) 5*sin(θ)}, we have:

[tex] 1: \ \ \ \ \ \setlength{\fboxsep}{4pt}\setlength{\fboxrule}{1pt}\frac{ \sin( \lim_{x \to 0} 5x)}{ \lim_{x \to 0} x} \,
\color{red} \blacktriangleleft \blacktriangleleft \fbox{\textsf{Not Equivalent}} \blacktriangleright \blacktriangleright \, \color{black} 5 \cdot ( \frac{ \sin( \lim_{x \to 0} x)}{ \lim_{x \to 0} x} ) [/tex]

The correct method to use is variable substitution {w = 5*x} so that {x = w/5},
and {(x → 0) ⇒ (w → 0)}:

[tex] 2: \ \ \ \ \ \lim_{x \to 0} \left(\frac{\sin(5x)}{x}\right) \ \ = \ \ \lim_{w \to 0} \left(\frac{\sin(w)}{w/5}\right) \ \ = \ \ 5 \cdot \lim_{w \to 0} \left(\frac{\sin(w)}{w}\right) \ \ = \ \ (5) \cdot (1) \ \ = \ \ \color{blue}(5) [/tex]


~~
 
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  • #26
p53ud0 dr34m5 said:
umm, you want a proof? let's see if i still remember the definition of a derivative

[tex]f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}[/tex]

so to prove:
[tex]\frac{d}{dx}sin(x)=cos(x)[/tex]

[tex]\lim_{h \to 0}\frac{sin(x+h)-sin(x)}{h}[/tex]

now, you need trig identities:
[tex]sin(a+b)=sin(a)cos(b)+sin(b)cos(b)[/tex]

[tex]\lim_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}[/tex]

[tex]\lim_{h \to 0}\frac{sin(x)(cos(h)-1)+sin(h)cos(x)}{h}[/tex]
[tex]\lim_{h \to 0}sin(x)*\lim_{h \to 0}\frac{cos(h)-1}{h}+\lim_{h \to 0}cos(x)*\lim_{h \to 0}\frac{sin(h)}{h}[/tex]

so:
[tex]sin(x)*0 + cos(x)*1 = cos(x)[/tex]
is that the proof you were looking for? i think its right; i don't really remember chapter two--chapter two was september. if so, then it is valid to use l'hopital's rule for his limit.

This standard (definition-using) proof prevents us from using l'Hôpital's rule for computing the limit [itex] \lim_{x\rightarrow 0} \mbox{sinc} \ x [/itex]...

Daniel.
 
  • #27
Okay, i think I finally understand now. I'd like to thank everyone for their help.
 

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

1. What does "lim -> inf" mean in this equation?

"lim -> inf" is a mathematical notation that represents the limit as a variable approaches infinity. In this equation, it indicates that we are looking at the behavior of x as it approaches infinity.

2. Why is the limit of x(sin(1/x)) equal to 1?

This result can be derived using the squeeze theorem, which states that if a function is sandwiched between two other functions that have the same limit, then the middle function also has the same limit. In this case, as x approaches infinity, sin(1/x) is bounded between -1 and 1, so x(sin(1/x)) is also bounded between -x and x. Since the limit of both -x and x as x approaches infinity is 0, the limit of x(sin(1/x)) must also be 0. Therefore, the limit is 1, as it is sandwiched between -x and x.

3. Can you provide a real-world example of the limit of x(sin(1/x))?

One real-world example that demonstrates the behavior of this limit is the speed of a car as it approaches infinity. As the distance traveled increases, the speed of the car will fluctuate between -1 and 1, but will ultimately approach 0 as the car reaches an infinite distance.

4. Are there any other ways to solve for the limit of x(sin(1/x))?

Yes, there are other methods to solve for the limit, such as using L'Hopital's rule or Taylor series expansion. However, the squeeze theorem is the most straightforward and efficient method for solving this particular limit.

5. What implications does this limit have in the field of mathematics?

This limit has several implications in mathematics, including its use in proving the convergence of series and determining the behavior of functions at infinity. It also demonstrates the concept of asymptotes, where a function approaches but never reaches a certain value. This limit is also a fundamental concept in calculus and is used in many real-world applications in physics and engineering.

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