What is my mistake in solving for this limit?

In summary: Yes, that is incorrect. ##\frac{d}{dn} \sin(1/n)= \cos(1/n) \cdot \frac d{dn}(1/n)##. If you carry out the last differentiation what do you get?The general rule when differentiating powers of ##x## is as follows:$$\frac{d}{dx}(x^n)=nx^{n-1}$$How can you write ##1/x## in the form of ##x^n##?If ##f(n)=n\sin(1/n)## then ##f'(n)=\sin(1/n)-\frac{\cos
  • #1
agnimusayoti
240
23
Homework Statement
Find the limit of the sequence as n approaching infinity.
Relevant Equations
lim n-->infty n sin (1/n)
I tried to substitution n = infinity so I got (infinity)*sin (1/infinity). I thought 1/infinity is approaching zero. So, sin (1/infinity) is same with sin (0). With these idea, my solution is lim n--> infinity n sin(1/n) = 0.

But, the answer book say that the answer is 1.

I tried another method with L'Hopital rule and came with 0/0 form.

So, here is my question:
1. Why my answer isn't valid? Isn't okay that I use sin (1/infinity) = 0?
2. Are there any methods beside the L'Hopital rule and "ln" strategic to find a limit?

Thanks a bunch pals.
 
Physics news on Phys.org
  • #2
Heuristically speaking

##\infty.\sin(1/\infty)=\infty.0## is an indeterminate form so you can't proceed like you did.

Rather, try to use l'Hospital's rule.
 
  • #3
Since you've tried the 0/0 form, then you probably got this:
$$\lim_{n\to\infty}\frac{\sin\frac 1n}{\frac 1n}$$
How did you apply l'Hôpital's rule? What's ##\frac{d}{dn}(\sin\frac 1n)##?
There're also some limits you might want to know:
limits-of-trigonometry-functions.jpg
 
  • #4
You might find it helpful to consider x=1/n tending to zero instead.
 
  • #5
Math_QED said:
Heuristically speaking

##\infty.\sin(1/\infty)=\infty.0## is an indeterminate form so you can't proceed like you did.

Rather, try to use l'Hospital's rule.
Yes, I have tried to use L'Hospital's rule and got this:
$$\lim_{n\rightarrow\infty} [{\sin (1/n)}-{\frac {cos (1/n)}{n}}]$$
Is that true?

Then, if I substitute ##\infty## to n I got
$$\sin (1/\infty) - \frac{cos (1/\infty)}{\infty}$$
$$\lim_{n\to\infty}n\sin(1/n)=(0)-\frac{\cos(0)}{\infty} $$

So far I don't see any mistake. But, if I did this, I came up with 0. This is not suitable with answer book, since I trust the answer book, rather then my answer. ..

Please give me insight. Thanks a lot
 
  • #6
archaic said:
Since you've tried the 0/0 form, then you probably got this:
$$\lim_{n\to\infty}\frac{\sin\frac 1n}{\frac 1n}$$
How did you apply l'Hôpital's rule? What's ##\frac{d}{dn}(\sin\frac 1n)##?
There're also some limits you might want to know:
View attachment 260823
$$\frac{d}{dn} \sin(1/n)=\sin(1/n)-\frac{cos(1/n)}{n}$$
Is that true?
 
  • #7
agnimusayoti said:
$$\frac{d}{dn} \sin(1/n)=\sin(1/n)-\frac{cos(1/n)}{n}$$
Is that true?

No. What you write is false. You need to apply the chain rule:

$$(f\circ g)'(x) = f'(g(x))g'(x)$$

Here ##g(x) = 1/x##, ##f(x) = \sin(x)##.
 
  • #8
Math_QED said:
No. What you write is false. You need to apply the chain rule:

$$(f\circ g)'(x) = f'(g(x))g'(x)$$

Here ##g(x) = 1/x##, ##f(x) = \sin(x)##.
Ups my mistake. What I write is for ##\frac{d}{dn} (n \sin (1/n))##

So, ##\frac{d}{dn} \sin (1/n)=-\frac{\cos(1/n)}{n}##
Is that any mistake? Thankss
 
  • #9
agnimusayoti said:
Ups my mistake. What I write is for ##\frac{d}{dn} (n \sin (1/n))##

So, ##\frac{d}{dn} \sin (1/n)=-\frac{\cos(1/n)}{n}##
Is that any mistake? Thankss

Still wrong. What is ##\frac{d}{dx}\frac{1}{x}##?
 
  • #10
agnimusayoti said:
Ups my mistake. What I write is for ##\frac{d}{dn} (n \sin (1/n))##
For this derivative, you need to use the product rule, which will give you two terms.
agnimusayoti said:
So, ##\frac{d}{dn} \sin (1/n)=-\frac{\cos(1/n)}{n}##
Is that any mistake? Thankss
Yes, that is incorrect. ##\frac{d}{dn} \sin (1/n)= \cos(1/n) \cdot \frac d{dn}(1/n)##. If you carry out the last differentiation what do you get?
 
  • #11
The general rule when differentiating powers of ##x## is as follows:
$$\frac{d}{dx}(x^n)=nx^{n-1}$$
How can you write ##1/x## in the form of ##x^n##?
 
  • #12
Ah, so I will get -1/n cos (1/n) right? Then, I substitute n = infnty. Is that so? Thankss
 
  • #13
agnimusayoti said:
Ah, so I will get -1/n cos (1/n) right? Then, I substitute n = infnty. Is that so? Thankss
You've got the minus sign right, but have forgotten to decrease the power.
$$\frac 1x=x^{-1}=x^n$$
 
  • #14
archaic said:
You've got the minus sign right, but have forgotten to decrease the power.
$$\frac 1x=x^{-1}=x^n$$
Yeaah, so ##\frac{d}{dn} \sin(1/n)=-1/n^2 \cos(1/n)## I hope it's true
 
Last edited by a moderator:
  • Like
Likes archaic
  • #15
agnimusayoti said:
Yeaah, so ##\frac{d}{dn} \sin(1/n)=-1/n^2 \cos(1/n)## I hope it's true
yeah that's it 👍
 
Last edited by a moderator:
  • #16
now go back to your limit and plug in the derivatives, something should cancel out.
 
  • #17
archaic said:
now go back to your limit and plug in the derivatives, something should cancel out.
If ##f(n)=n\sin(1/n)## then ##f'(n)=\sin(1/n)-\frac{\cos(1/n)}{n}##. Is that true?
Then by L'Hospital's rule; $$\lim_{n\to\infty} n\sin(1/n)=\lim_{n\to\infty} \sin(1/n) - \lim_{n\to\infty} \frac{\cos(1/n)}{n}$$
Can substitute ##n=\infty## to the function so I get:
$$\lim_{n\to\infty} n\sin(1/n)=\sin(1/\infty) - \frac{\cos(1/\infty)}{\infty}$$
$$\lim_{n\to\infty} n\sin(1/n)=\sin(0) - \frac{\cos(0)}{\infty}$$
$$\lim_{n\to\infty} n\sin(1/n)=0 - 0=0$$

Is it true? If that's true, the answer is still different from the key. Hiks. I am not confidence with my answer, though.
 
  • #18
agnimusayoti said:
Then by L'Hospital's rule;
No, you are not applying it correctly. You must first express the function you want the limit of as the ratio of two functions then differentiate each.

I still think you will find it easier if you replace n with 1/x and consider the limit as x tends to zero.
 
  • #19
Capture.PNG

There's also the requirement that ##\lim_{x\to a}\frac{f'(x)}{g'(x)}## exists. If it doesn't, you can either apply the rule again, if possible, or try other methods.

That's the theorem. You need to transform your limit into either of those forms before applying it.
 
  • #20
Gee! I think L'Hospital's rule valid for any form. Thanks for
archaic said:
View attachment 260875
There's also the requirement that ##\lim_{x\to a}\frac{f'(x)}{g'(x)}## exists. If it doesn't, you can either apply the rule again, if possible, or try other methods.

That's the theorem. You need to transform your limit into either of those forms before applying it.
So, like haruspex explain, I assume that ##x=1/n## then I will get ##f(n)=n sin(1/n)## equivalent with ##f(x)=\sin(x)/x##

Then, I apply the L'Hospital rule.So, what's wrong is my concept about L'Hospital rule. Thanks everyone. I will correct my answer.
 
  • #21
After I made correction, there's no need to apply L'Hospital because
$$x=1/n$$
so,
$$\lim_{n\to\infty} n \sin(1/n) \equiv \lim_{x\to0} \frac {sin x}{x}$$
By the theorem:
$$\lim_{x\to0} \frac{\sin x}{x}=1$$

Is that true, shifu?
 
  • Like
Likes archaic
  • #22
Using l'Hôpital's rule:
$$n\sin\frac 1n=\frac{1}{\frac 1n}\sin\frac 1n=\frac{\sin\frac 1n}{\frac 1n}\\
\lim_{n\to\infty}n\sin\frac 1n=\lim_{n\to\infty}\frac{\sin\frac 1n}{\frac 1n}=\lim_{n\to\infty}\frac{-\frac{1}{n^2}\cos\frac 1n}{-\frac{1}{n^2}}=\lim_{n\to\infty}\cos\frac 1n=1$$
Another way is with the substitution you made:
$$\lim_{n\to\infty}n\sin\frac 1n=\lim_{n\to\infty}\frac{\sin\frac 1n}{\frac 1n}=\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\cos x}{1}=1$$
Or, also, directly using the theorem as you did.
 
  • Like
Likes agnimusayoti
  • #23
agnimusayoti said:
After I made correction, there's no need to apply L'Hospital because
$$x=1/n$$
so,
$$\lim_{n\to\infty} n \sin(1/n) \equiv \lim_{x\to0} \frac {sin x}{x}$$
By the theorem:
$$\lim_{x\to0} \frac{\sin x}{x}=1$$

Is that true, shifu?
It is not clear whether you are permitted to use that result or, alternatively, you are effectively being asked to prove it.
 
Last edited:
  • #24
archaic said:
Using l'Hôpital's rule:
$$n\sin\frac 1n=\frac{1}{\frac 1n}\sin\frac 1n=\frac{\sin\frac 1n}{\frac 1n}\\
\lim_{n\to\infty}n\sin\frac 1n=\lim_{n\to\infty}\frac{\sin\frac 1n}{\frac 1n}=\lim_{n\to\infty}\frac{-\frac{1}{n^2}\cos\frac 1n}{-\frac{1}{n^2}}=\lim_{n\to\infty}\cos\frac 1n=1$$
Another way is with the substitution you made:
$$\lim_{n\to\infty}n\sin\frac 1n=\lim_{n\to\infty}\frac{\sin\frac 1n}{\frac 1n}=\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\cos x}{1}=1$$
Or, also, directly using the theorem as you did.
Yes! Thank you. In this case I can use the theorem .
 
  • Like
Likes archaic

Related to What is my mistake in solving for this limit?

1. What is a limit in mathematics?

A limit in mathematics is a fundamental concept that describes the behavior of a function as its input approaches a certain value. It is used to determine the value that a function approaches as the input gets closer and closer to a specific value.

2. How do I know if I have made a mistake in solving for a limit?

If you have made a mistake in solving for a limit, you may notice that the answer you have obtained does not match the expected result. Additionally, you may encounter an undefined or indeterminate form in your calculation, such as 0/0 or ∞/∞.

3. What are the common mistakes made when solving for a limit?

Some common mistakes when solving for a limit include forgetting to consider the left and right-hand limits, using incorrect algebraic manipulations, and not simplifying the expression enough before taking the limit.

4. How can I avoid making mistakes when solving for a limit?

To avoid making mistakes when solving for a limit, it is important to carefully follow the steps and rules for evaluating limits. This includes understanding the properties of limits, using the correct techniques for different types of limits, and double-checking your work for any errors.

5. What resources can I use to help me solve limits correctly?

There are many resources available to help you solve limits correctly, including textbooks, online tutorials, and practice problems. You can also seek help from a math tutor or consult with your classmates or professor for additional guidance and support.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
802
  • Calculus and Beyond Homework Help
Replies
5
Views
462
  • Calculus and Beyond Homework Help
Replies
2
Views
929
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
789
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Back
Top