Solving Limits with Trig Functions

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Homework Help Overview

The discussion revolves around solving limits involving trigonometric functions, specifically the limit as x approaches zero for the expression (x + xcosx) / (sinxcosx). Participants are exploring the application of known limits and properties of trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the manipulation of the limit expression and the relevance of the limit limθ→0 sinθ/θ = 1. There are questions about the behavior of cosx as x approaches zero and the validity of the expression x/sinx equating to 1.

Discussion Status

The discussion is active, with participants offering insights and questioning assumptions. Some guidance has been provided regarding the limit properties, but there is no explicit consensus on the interpretation of the limit or the steps to take next.

Contextual Notes

Participants are navigating the nuances of limit laws and the implications of trigonometric identities, with some expressing skepticism about certain steps in the reasoning process. There is an acknowledgment of grading criteria that may affect how limits are approached.

TommG
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I am having trouble solving limits with trig functions.

limx→0 x+xcosx/sinxcosx

useful equation limθ→0 sinθ/θ = 1

my attempt

x+xcosx/sinxcosx

x/sinxcosx + xcosx/sinxcosx

x/sinx*1/cosx + x/sinx this is where I am getting stuck
 
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the answer is right in front of you.

As x approaches zero then what is the value of cosx?

and you have the sinx/x as x approaches zero so everything is there for you
 
jedishrfu said:
the answer is right in front of you.

As x approaches zero then what is the value of cosx?

and you have the sinx/x as x approaches zero so everything is there for you


Does x/sinx = 1 also?

I thought sinx had to be in the numerator for it to = 1
 
TommG said:
Does x/sinx = 1 also?

I thought sinx had to be in the numerator for it to = 1

Technically, you're right. But note that if ##x\neq 0##, then ##\frac{x}{\sin x}=\frac{1}{\frac{\sin x}{x}}##, and you have something to which you can apply your limit law for ratios.

So I guess technically you're not right (with regard to the value of the limit), but it's good to be skeptical. I had some students a year or so ago that got marked down (per a grading rubric that was not mine) for omitting the step that I showed.
 

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