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

[camboguy]

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.

 

[Dick]

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.

 

[camboguy]

What do you mean by its "not enough of a squeeze to force it to a lim"?

 

[Dick]

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.