Demonstrating Limit Equalities for x→0

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around demonstrating the limit equalities as x approaches 0 for the expressions involving sine and cosine functions. The focus is on various methods to prove these limits, including algebraic manipulation, the Squeeze Theorem, L'Hôpital's Rule, and geometric interpretations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests a solid demonstration of the limits: lim_{x→0} (sin x)/x = 1 and lim_{x→0} (1-cos x)/x = 0.
  • Another participant suggests using the Squeeze Theorem along with trigonometric inequalities: sin{x}cos{x} ≤ x ≤ tan{x}.
  • A participant questions whether L'Hôpital's Rule is an adequate method for demonstrating these limits.
  • Another participant agrees that L'Hôpital's Rule is valid but notes that these limits are often introduced before students learn derivatives or L'Hôpital's Rule.
  • One suggestion includes using Taylor expansions around x=0 to demonstrate the limits.
  • A participant mentions using geometric interpretations involving triangles and areas to derive the inequalities leading to the limits, emphasizing a non-calculus approach.

Areas of Agreement / Disagreement

Participants express differing opinions on the best methods to demonstrate the limits, with some favoring algebraic approaches and others supporting L'Hôpital's Rule or geometric interpretations. No consensus on a single preferred method is reached.

Contextual Notes

Some methods proposed depend on prior knowledge of calculus concepts, while others suggest non-calculus approaches. The discussion reflects varying levels of familiarity with the techniques involved.

Werg22
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I need a solid demonstration of those two equalities

[tex]{_\lim{x}{\rightarrow} 0}\frac{\sin {x}}{x} = 1[/tex]

[tex]{_\lim{x}{\rightarrow} 0}\frac{1-\cos {x}}{x} = 0[/tex]

How to do so?
 
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Use the following fact and do some algebra / trig. to rearrange, then use the Squeeze Theorem:

[tex]\sin{x}\cos{x}\leq x\leq\tan{x}[/tex]

For the second, multiply the numerator and denominator by 1+cos(x) and use the results of the first when simplifying.
 
is l'hopital's rule not a good enough demonstration?
 
Yeah, that works. The only reason I did so otherwise is because generally these limits are introduced at the beginning of an introductory calculus course, before students have had derivatives or L'Hopital's Rule. At least, that is how it was for me.
 
Or try the Taylor expansions about x=0.
 
triangles

It is usually done without calculus. Just because it is used to define the derivatives of certain functions. If you compare the areas in the graph you can come up with the inequalities and use the squeeze theorem. Probably works out to the same formula above, but with a little more motivation.:bugeye:
 
sorry, forgot the picture. Sorry it's not very "pretty"
 

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