Need Help with My Limit Homework

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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:

[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. :)
 
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..
 
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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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?

[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??
 
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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?
 
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.