• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Evaluate the following limits if it exist

  • Thread starter haha1234
  • Start date
1. The problem statement, all variables and given/known data

evaluate the following limits if it exist
lim x sin1/x and limit x sin 1/x
x→0 x→∞
2. Relevant equations



3. The attempt at a solution
Someone have told me that I should let t=1/x and rewrite the limits.However, once I rewrite the limit, I still cannot evaluate the limits.
So,how can I find the limits?
 

CAF123

Gold Member
2,882
87
First rewrite the expression as $$\lim_{x \rightarrow 0} \frac{\sin\left(1/x\right)}{1/x}$$ and then the reason for introducing the variable ##t## becomes more obvious.
 

HallsofIvy

Science Advisor
41,625
822
You should have learned the limit [tex]\lim_{\theta\to 0}\frac{sin(\theta)}{\theta}[/tex] in first semester Calculus. If you do not remember it, look in a Calculus text in the section on the derivative of sin(x).
 
32,349
4,135
1. The problem statement, all variables and given/known data

evaluate the following limits if it exist
lim x sin1/x and limit x sin 1/x
x→0 x→∞
2. Relevant equations



3. The attempt at a solution
Someone have told me that I should let t=1/x and rewrite the limits.However, once I rewrite the limit, I still cannot evaluate the limits.
So,how can I find the limits?
The OP's problem statement wasn't very clear, but there are two problems here.

$$\lim_{x \to 0} x sin(1/x)$$
$$\lim_{x \to \infty} x sin(1/x)$$

The hint applies to the second problem.
 
You should have learned the limit [tex]\lim_{\theta\to 0}\frac{sin(\theta)}{\theta}[/tex] in first semester Calculus. If you do not remember it, look in a Calculus text in the section on the derivative of sin(x).
I have used this method,but I still cannot find correct answer that I got the answer is 1.
 
First rewrite the expression as $$\lim_{x \rightarrow 0} \frac{\sin\left(1/x\right)}{1/x}$$ and then the reason for introducing the variable ##t## becomes more obvious.
Why limx→0xsin(1/x)is the same as limx→0sin(1/x)1/x ?:confused::shy:
 

Office_Shredder

Staff Emeritus
Science Advisor
Gold Member
3,734
98
haha, the only thing they did to rewrite the limit like that is
[tex] x = \frac{1}{1/x} [/tex]
 

Want to reply to this thread?

"Evaluate the following limits if it exist" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top