Limits of fractions of polynomials and trig functions

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Homework Help Overview

The discussion revolves around evaluating limits of fractions involving polynomials and trigonometric functions. Participants are examining specific limits and the methods to approach them.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss plugging in values directly into the limit expressions and express confusion when encountering indeterminate forms. There is an exploration of simplifying expressions, particularly in relation to square roots and polynomial factors.

Discussion Status

Some participants have provided guidance on simplifying expressions and questioning the correctness of the limit statements. There is an ongoing exploration of assumptions regarding the limit points, with no explicit consensus reached on the nature of the problems presented.

Contextual Notes

There is mention of potential typos in the limit statements provided by the teacher, leading to discussions about the validity of the limits as stated. Participants are navigating through these uncertainties without resolving them definitively.

carbz
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I have two...

Homework Statement


The the limit


Homework Equations


\lim_{x \rightarrow 1} \frac{1-cosx}{x^2}


The Attempt at a Solution


I figured to just plug in 1, but I wanted to make sure...

Homework Statement


Find the limit


Homework Equations


\lim_{x \rightarrow 3} \frac{\sqrt{x^2-6x+9}}{x-3}


The Attempt at a Solution


I plugged in the 3, and got 3/0, then I got lost...
 
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In 2, did you try simplifying the numerator? (What are the roots of the polynomial?)
 
Yes, I tried doing that.

(x-3)(x-3)

However, I forgot how to get rid of that radical. Squaring wouldn't work, so I have no idea.


Also, no thoughts on the first one?
 
What is a short hand expression for (x-3)(x-3)?
 
(x-3}^2. Oh yeah, so that takes away the square root, and after everything, it leaves 0. thank you.
 
In the first one, are you sure the problem isn't x->0 instead of x->1 ?
 
it is 1, not 0.
 
Well, then your book has a typo..
 
it's not from my book, it was my teacher.
 
  • #10
Well, it's 99% certain that your teacher meant to write 0 instead of 1. With 1, it's trivial, since both the numerator and denominator are finite, nonzero constants in that limit.
 
  • #11
Then he either blundered, or tried to fool you.

Your function is defined&continuous on all values of x except x=0.

Your original approach is perfetly valid in the case of x=1.
 
  • #12
allright, thankyou.
 

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