Need Help with My Limit Homework

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The discussion centers on finding the limit as x approaches zero from the right for the expression x * |1/x|, which is initially misunderstood. Participants clarify that the absolute value can be simplified, leading to the correct interpretation of the limit. The ceiling function is identified as the intended notation, prompting further analysis using inequalities. The Squeeze Theorem is suggested as a method to determine the limit, emphasizing the importance of understanding the behavior of the expressions near zero. Ultimately, the correct limit is confirmed to be 1.
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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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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:

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

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

You know that:

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

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. :)
 
VietDao29 said:
Is this your limit:

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

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

You know that:

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

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..
 
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?

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

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

\frac{1}{x} \leq \left\lceil \frac{1}{x} \right\rceil \leq \frac{1}{x} + 1

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

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

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

\frac{1}{x} \leq \left\lceil \frac{1}{x} \right\rceil \leq \frac{1}{x} + 1

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??
 
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what is the answer?
 
emyt said:
what is the answer?

Use the Squeeze Theorem on the inequality in post #4.
 
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
 
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