Evaluate the following limits if it exist

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Do you see the difference?In summary, the conversation discusses evaluating the limits lim x sin(1/x) as x approaches 0 and x sin(1/x) as x approaches infinity. The suggested method is to rewrite the expression as sin(1/x)/(1/x) and use the limit rule \lim_{\theta\to 0}\frac{sin(\theta)}{\theta}. However, the OP is still struggling to find the correct answer. The confusion lies in rewriting the expression as x = \frac{1}{1/x}, which is incorrect.
  • #1
haha1234
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Homework Statement



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

Homework Equations





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?
 
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  • #2
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.
 
  • #3
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).
 
  • #4
haha1234 said:

Homework Statement



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

Homework Equations





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.
 
  • #5
HallsofIvy said:
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.
 
  • #6
CAF123 said:
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:
 
  • #7
haha, the only thing they did to rewrite the limit like that is
[tex] x = \frac{1}{1/x} [/tex]
 
  • #8
haha1234 said:
Why limx→0xsin(1/x)is the same as limx→0sin(1/x)1/x ?:confused::shy:
It isn't the same. What you wrote is sin(1/x) * 1/x. What it should be is sin(1/x)/(1/x).
 

1. What does it mean to evaluate a limit?

Evaluating a limit means finding the value that a function approaches as its input approaches a certain value.

2. How do I know if a limit exists?

A limit exists if the value of the function approaches a single value as the input approaches the given value. This can be determined by evaluating the left and right-hand limits at the given value and seeing if they are equal.

3. What does it mean if a limit does not exist?

If a limit does not exist, it means that the function does not approach a single value as the input approaches the given value. This could be due to a jump or a discontinuity in the function.

4. Can a limit be infinite?

Yes, a limit can be infinite if the function approaches positive or negative infinity as the input approaches the given value. This can happen if the function has a vertical asymptote.

5. How do I evaluate a limit algebraically?

To evaluate a limit algebraically, you can use techniques such as factoring, simplifying, and canceling common terms. You can also use L'Hopital's rule or the squeeze theorem if the function is indeterminate at the given value.

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