Need Help with My Limit Homework

In summary: Remember that x is not zero; the limit only examines the behavior of these expressions in arbitrarily small neighborhoods of 0. Use the Squeeze Theorem to imply the limit of the central expression.right, thanks!
  • #1
ModalNekat
3
0

Homework Statement


Limit x approached zero from right which is X * [|1/X|]

Homework Equations


Since i Don't know how to draw the graphs, then i don't have an equations


The Attempt at a Solution


I've tried to drew the graphic, but from the graphs, I've concluded that lim x aprroached zero from right [|1/x|] does not exist because 1/0 = undefined.

Actually, the answer is 1, but still i don't know how to solve this.
So, guys, please help me up!Thanks
 
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  • #2
ModalNekat said:

Homework Statement


Limit x approached zero from right which is X * [|1/X|]

Homework Equations


Since i Don't know how to draw the graphs, then i don't have an equations


The Attempt at a Solution


I've tried to drew the graphic, but from the graphs, I've concluded that lim x aprroached zero from right [|1/x|] does not exist because 1/0 = undefined.

Actually, the answer is 1, but still i don't know how to solve this.
So, guys, please help me up!Thanks

Is this your limit:

[tex]\lim_{x \rightarrow 0 ^ {+}} \left( x \left| \frac{1}{x} \right| \right)[/tex]?

Well, the first thing when dealing with absolute value, is to take out all absolute signs.

You know that:

[tex]|a| = \left\{ \begin{array}{ll} a & \mbox{, if } a \geq 0 \\ -a & \mbox{, if } a < 0 \end{array} \right.[/tex]

So, when x tends to 0+, (i.e it tends to 0 from the right), is 1/x positive or negative? Can you break absolute signs?

After breaking (taking out) all the absolute signs, you should be arriving to the final answer shortly. :)
 
  • #3
VietDao29 said:
Is this your limit:

[tex]\lim_{x \rightarrow 0 ^ {+}} \left( x \left| \frac{1}{x} \right| \right)[/tex]?

Well, the first thing when dealing with absolute value, is to take out all absolute signs.

You know that:

[tex]|a| = \left\{ \begin{array}{ll} a & \mbox{, if } a \geq 0 \\ -a & \mbox{, if } a < 0 \end{array} \right.[/tex]

So, when x tends to 0+, (i.e it tends to 0 from the right), is 1/x positive or negative? Can you break absolute signs?

After breaking (taking out) all the absolute signs, you should be arriving to the final answer shortly. :)
Whoops, I'm sorry for misunderstanding, looks like [| |] is not dealing with Greatest INteger Function. Then, what i mean with [|x|] is [[ x ]], sorry..
 
  • #4
ModalNekat said:
Whoops, I'm sorry for misunderstanding, looks like [| |] is not dealing with Greatest INteger Function. Then, what i mean with [|x|] is [[ x ]], sorry..

You mean the http://en.wikipedia.org/wiki/Floor_and_ceiling_functions" , right? Or are you referring to the Floor Function?

[tex]\lim_{x \rightarrow 0 ^ {+}} \left( x \left\lceil \frac{1}{x} \right\rceil \right)[/tex]

If you really mean the Ceiling Function, then from its definition, we can derive to the following inequality:

[tex]\frac{1}{x} \leq \left\lceil \frac{1}{x} \right\rceil \leq \frac{1}{x} + 1[/tex]

Can you find the limit of the expression based on this inequality?
 
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  • #5
I've always known [|x|] to be the greatest integer function, or floor function.
Are you saying it actually looks like [[x]] instead?
 
  • #6
VietDao29 said:
You mean the http://en.wikipedia.org/wiki/Floor_and_ceiling_functions" , right? Or are you referring to the Floor Function?

[tex]\lim_{x \rightarrow 0 ^ {+}} \left( x \left\lceil \frac{1}{x} \right\rceil \right)[/tex]

If you really mean the Ceiling Function, then from its definition, we can derive to the following inequality:

[tex]\frac{1}{x} \leq \left\lceil \frac{1}{x} \right\rceil \leq \frac{1}{x} + 1[/tex]

Can you find the limit of the expression based on this inequality?
Yeah, i mean it's the ceiling function. Thanks a lot, right now it could be derived.Actually i got this problems from Purcell's Calculus Book, do u have one on you??
 
Last edited by a moderator:
  • #7
what is the answer?
 
  • #8
emyt said:
what is the answer?

Use the Squeeze Theorem on the inequality in post #4.
 
  • #9
the interval between 1/x and 1/x + 1 is fairly large though?
 
  • #10
emyt said:
the interval between 1/x and 1/x + 1 is fairly large though?

Multiply through by x, then take the limit of each expression. Remember that x is not zero; the limit only examines the behavior of these expressions in arbitrarily small neighborhoods of 0. Use the Squeeze Theorem to imply the limit of the central expression.
 
  • #11
right, thanks
 

Related to Need Help with My Limit Homework

1. What is a limit in mathematics?

A limit in mathematics is a fundamental concept that describes the behavior of a function as the 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 particular point.

2. Why do we need to learn about limits?

Limits are essential in mathematics because they help us understand the behavior of functions and their graphs. They are also used to define important concepts such as derivatives and integrals, which are crucial in calculus and other branches of mathematics.

3. How do I solve limit problems?

The process of solving limit problems involves evaluating the function at different points, both numerically and graphically, and observing the behavior of the function as the input approaches the desired value. Algebraic techniques, such as factoring and simplifying, are also useful in solving limit problems.

4. What are the common types of limit problems?

The most common types of limit problems include evaluating limits at infinity, evaluating limits involving trigonometric functions, and evaluating limits involving rational functions. Other types may involve piecewise-defined functions, exponential functions, or logarithmic functions.

5. How can I check if my limit solution is correct?

To check if your limit solution is correct, you can use a graphing calculator or an online limit calculator to compare your answer with the expected result. You can also verify your solution algebraically by plugging in the value of the limit into the original function and simplifying it to see if it matches your answer.

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