Finding the limit of this expression

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression \(\frac{8x^3+x^2-5x}{3x^4-5x^2+2}\) as \(x\) approaches 1, focusing on the behavior of the numerator and denominator near this point. Additionally, there is a mention of another limit involving the greatest integer function as \(x\) approaches 0.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting values into the limit expression and analyze the behavior of the numerator and denominator as \(x\) approaches 1. Questions arise regarding the implications of the signs of the numerator and denominator near this limit. There is also an exploration of a second limit involving the greatest integer function, with predictions about its outcome.

Discussion Status

The discussion is ongoing, with participants providing insights into the behavior of the limit as \(x\) approaches 1. Some participants suggest that the limit does not exist due to differing left-side and right-side limits, while others are exploring a new limit involving the greatest integer function, indicating a productive exchange of ideas.

Contextual Notes

Participants note the challenge of evaluating limits that result in indeterminate forms and the implications of approaching from different sides. There is also mention of using external tools like WolframAlpha to verify results, which introduces potential confusion regarding the limits being evaluated.

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Homework Statement



The limit of [tex]\frac{8x^3+x^2-5x}{3x^4-5x^2+2}[/tex] as x goes to 1

Homework Equations



Evaluate this limit algebraically.

The Attempt at a Solution



Tried using 1 instead of x. The denominator becomes 0.
Divided denominator by (x-1) but couldn't continue.
 
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When x is near 1, the numerator will be near 4, which is a positive number.

When x is near 1, but larger than 1, the denominator is near zero and is positive. When x is near 1, but less than 1, the denominator is near zero and is negative.

Can you conclude anything from this information, about the limit of your rational function as x approaches 1?
 
Mark44 said:
When x is near 1, the numerator will be near 4, which is a positive number.

When x is near 1, but larger than 1, the denominator is near zero and is positive. When x is near 1, but less than 1, the denominator is near zero and is negative.

Can you conclude anything from this information, about the limit of your rational function as x approaches 1?

does that mean limit does not exist?
 
Yes, but can you clarify how you reached that conclusion?
 
Mark44 said:
Yes, but can you clarify how you reached that conclusion?

at first thank for your help
i believe so.

while x is approaching 1 from + side the denominator gets closer to 0 which makes the limit positive infinity as x approaches 1+

while x is approaching 1 from - side, the limit becomes negative infinity.

one side of x=1 goes to -infinity the other +infinity so i conclude that the limit does not exist.

now i am working on this limit:

greatest integer(sinx) / greatest integer(x)
x approaches to 0

i predict that result will be 0 but i am not sure
 
[tex]\left\lfloor[/tex]
turutk said:
at first thank for your help
i believe so.

while x is approaching 1 from + side the denominator gets closer to 0 which makes the limit positive infinity as x approaches 1+

while x is approaching 1 from - side, the limit becomes negative infinity.

one side of x=1 goes to -infinity the other +infinity so i conclude that the limit does not exist.
OK, good. Since the left-side limit is different from the right-side limit, the limit itself doesn't exist.
turutk said:
now i am working on this limit:

greatest integer(sinx) / greatest integer(x)
x approaches to 0

i predict that result will be 0 but i am not sure
I don't think it's zero, but I'm not sure, either. I would make a table of values of x and sin(x) values for x near zero on either side. Keep in mind that when -pi/2 < sin(x) < 0,
[tex]\left \lfloor{sin(x)}\right \rfloor = -1[/tex]
 
Mark44 said:
[tex]\left\lfloor[/tex]OK, good. Since the left-side limit is different from the right-side limit, the limit itself doesn't exist.

I don't think it's zero, but I'm not sure, either. I would make a table of values of x and sin(x) values for x near zero on either side. Keep in mind that when -pi/2 < sin(x) < 0,
[tex]\left \lfloor{sin(x)}\right \rfloor = -1[/tex]

turns out it is zero:
http://www.wolframalpha.com/input/?...nx))/greatest+integer(x)+as+x+approaches+to+2

i still cannot explain
 

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