How can the squeeze theorem be used to find g(x) and h(x) for a given limit?

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Homework Help Overview

The discussion revolves around the limit lim(x→0) x^4 cos(5/x) and the application of the squeeze theorem to identify bounding functions g(x) and h(x) for the function f(x) = x^4 cos(5/x).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the undefined nature of the limit at x=0 and the implications of substituting values directly into the function. They explore the behavior of the cosine function and its bounds, questioning how to apply these bounds to find g(x) and h(x).

Discussion Status

Some participants have suggested using the known bounds of the cosine function (-1 ≤ cos(θ) ≤ 1) to establish the bounding functions for f(x). There is an ongoing exploration of how to connect these ideas to formulate g(x) and h(x), with some participants expressing uncertainty about the next steps.

Contextual Notes

Participants note the challenge of working with a function that is undefined at a certain point and the need to understand the behavior of the function around that point without direct substitution.

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Homework Statement


Given the limit
lim(x→0) x^4 cos(5/x)
use the squeeze theorem to find g(x) and h(x) given,
f(x)=x^4 cos(5/x)

g(x)≤f(x)≤h(x)

The Attempt at a Solution


well the limit with x=0 substituted in would mean the limit is undefined as,
0^4 cos(5/0)
(5/0) can not occur but due to the function being trigonometric the usual use of 0^- and 0^+ can not be used, i found as -1≤cos θ≤1

i have no idea how to find the g(x) and h(x) with only an f(x) function, can anyone please talk me through how to use the squeeze theorem for this problem?
 
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steve snash said:

Homework Statement


Given the limit
lim(x→0) x^4 cos(5/x)
use the squeeze theorem to find g(x) and h(x) given,
f(x)=x^4 cos(5/x)

g(x)≤f(x)≤h(x)

The Attempt at a Solution


well the limit with x=0 substituted in would mean the limit is undefined as,
0^4 cos(5/0)
The whole idea of limits is to be able to determine the behavior of a function that is undefined at some point. Except in the simplest cases, where a limit is unnecessary anyway, you NEVER just substitute the limiting x value into the function.

Just because x4 cos(5/x) is undefined at 0 doesn't mean that the limit doesn't exist. For example (sin x)/x is undefined at x = 0, yet the limit of this function as x approaches 0 does exist, and is in fact equal to 1.
steve snash said:
(5/0) can not occur but due to the function being trigonometric the usual use of 0^- and 0^+ can not be used, i found as -1≤cos θ≤1
You have the seed of an idea here, since -1 <= cos(whatever) <= 1. How can you apply this idea to your problem to find two functions that bound x4 cos(5/x)?
steve snash said:
i have no idea how to find the g(x) and h(x) with only an f(x) function, can anyone please talk me through how to use the squeeze theorem for this problem?
 
Mark44 said:
You have the seed of an idea here, since -1 <= cos(whatever) <= 1. How can you apply this idea to your problem to find two functions that bound x4 cos(5/x)?

yes i know that i can use this to find the lower and upper (gx and hx) but my problem is how do i use this (-1<=cos theta<=1) to find the gx and hx??

can someone walk me through how to use this knowledge, where do i start, i need to know this for study purposes.
 
Come on - think!
You have f(x) = x4cos(5/x), and you know that -1 <= cos(whatever) <= 1. Can't you connect the dots?
 
yeah i worked it out =p man I am a dummy thanks mark
 

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