Finding a Trig Limit by hand, no L'Hopitals

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SUMMARY

The limit of the expression \(\lim_{x\rightarrow 0}\frac{\cos^2x-1}{2x\sin x}\) evaluates to -0.5 without using L'Hôpital's rule. By applying the identity \(1 = \cos^2(x) + \sin^2(x)\) and substituting it into the limit, simplification leads to a known limit that can be utilized. The key limits used in this discussion are \(\lim_{x\rightarrow 0}\frac{1-\cos x}{x}=0\) and \(\lim_{x\rightarrow 0}\frac{\sin x}{x}=1\).

PREREQUISITES
  • Understanding of trigonometric identities, specifically \(1 = \cos^2(x) + \sin^2(x)\)
  • Familiarity with limits in calculus
  • Knowledge of the limit properties \(\lim_{x\rightarrow 0}\frac{1-\cos x}{x}=0\) and \(\lim_{x\rightarrow 0}\frac{\sin x}{x}=1\)
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study trigonometric identities and their applications in limits
  • Practice solving limits without L'Hôpital's rule
  • Explore the concept of continuity and its relation to limits
  • Learn about Taylor series expansions for trigonometric functions
USEFUL FOR

Students in calculus courses, particularly those struggling with trigonometric limits, and educators seeking to enhance their teaching methods in limit evaluation.

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Homework Statement



[itex]\stackrel{lim}{x\rightarrow 0}\frac{cos^2x-1}{2xsinx}[/itex]


Homework Equations



[itex]\stackrel{lim}{x\rightarrow 0}\frac{1-cosx}{x}=0[/itex]

[itex]\stackrel{lim}{x\rightarrow 0}\frac{sinx}{x}=1[/itex]

The Attempt at a Solution



I found this problem online (and can't remember where). It was in a limits section, so it can be solved without using l'hospital's rule. I gave it to my class today, and then we all got stuck. We know the answer is -.5 but can't get it algebraically. Any advice?
 
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Remember that 1 = cos^2(x) + sin^2(x).

Now substitute and simplify and you can use one of your known limits.
 
Thank you! I got thrown into this class half way through the year, and while I have re-learned the calc, my trig is horrible. Thank you!
 

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