What is the Limit of a Squeezed Function with Shifting Center?

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Homework Help Overview

The discussion revolves around finding the limit of the function (3/4x) cos(4/(x+2)) as x approaches -2. The original poster expresses confusion regarding the application of the squeeze theorem, particularly due to the shifting center and the behavior of the cosine function near this point.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to analyze the limit by considering the bounds of the function, noting the oscillatory nature of the cosine term and its implications for the limit. They question whether the limit exists or approaches zero based on their observations of the bounding functions.

Discussion Status

Some participants have provided insights regarding the behavior of the function near the limit point, suggesting that the oscillation does not converge to a single value. There is an ongoing exploration of the implications of the squeeze theorem in this context, with differing interpretations of the limit's existence.

Contextual Notes

The original poster mentions the challenge of analyzing the limit without graphing tools, which adds to their uncertainty regarding the behavior of the functions involved.

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Homework Statement



find the lim of:

lim X-> -2 (3/4x) cos (4/(x+2))



The Attempt at a Solution



The x-> -2 threw me off, because the center changed.

im having trouble understanding the function of the squeeze therm here.
i understand that the bound is (-3/4x) and (3/4x). i also understand that the graph of the cos is shifted to the left by 2 units and so as x-> -2 it is centered at -2.

i was confused when i took the lim of (-3/4x) and got (3/8), and for (3/4x) i got (-3/8).
since the lim don't equal i am going assuming the lim DNE. But on the other hand when i graph (-3/4x) and (3/4x) they make and X graph centered on the origin. Also the graph of the cos(4/(2+x)) oscillates wildly at -2 but cos(4/(2+x)) is also squeezed at the origin between (-3/4x) and (3/4x).

so would the lim be zero? or would it be DNE? i really think its zero since the bounds of (-3/4x) and (3/4x) squeeze cos (4/(x+2)), but i don't understand how i am supposed to do it without a graphing calculator or graphing it by hand to see where the functions are being squeezed.

Thank you in advance.
 
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Your first guess, DNE, is the correct one. And you are also right that very near x=-2 the function oscillates infinitely between about -3/8 and +3/8 (since 4/(2+x) goes to +/-infinity it goes through an infinite number of cycles of cos). That's not enough of a squeeze to force it to a limit. You don't really need a graphing calculator, to make a rough picture of this in your head.
 
What do you mean by its "not enough of a squeeze to force it to a lim"?
 
I mean you are only forcing the oscillation to be between -3/8 and 3/8. There's a lot of numbers in that range. The limit doesn't exist.
 

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