Limit with trigonometric functions

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SUMMARY

The limit of Tan(x)/(1-Cos(x)) as x approaches π from the left is 0, confirming that the statement is false. The limit of Sec^2(x)/Sin(x) also approaches infinity, but this does not equate to the first limit. The application of L'Hôpital's Rule is unnecessary as the original limit does not present indeterminate forms. Therefore, the conclusion drawn from the initial evaluation is valid and accurate.

PREREQUISITES
  • Understanding of trigonometric functions, specifically Tan(x), Cos(x), and Sec(x).
  • Knowledge of limits and their properties in calculus.
  • Familiarity with L'Hôpital's Rule and its application to indeterminate forms.
  • Ability to evaluate limits analytically without reliance on numerical substitution.
NEXT STEPS
  • Study the properties of limits involving trigonometric functions.
  • Learn about L'Hôpital's Rule and its appropriate applications.
  • Explore advanced limit techniques, such as Taylor series expansions.
  • Practice evaluating limits that yield indeterminate forms.
USEFUL FOR

Students studying calculus, particularly those focusing on limits and trigonometric functions, as well as educators seeking to clarify common misconceptions in limit evaluation.

Jules18
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Homework Statement



True or false?

lim Tanx/(1-Cosx) = lim Sec2x/Sinx = Infinity

(limits are as x approaches \pi from the left)

The Attempt at a Solution



I tried just plugging \pi into x in the first limit, and I ended up getting 0/2, which exists but is just 0. So I said false. Am I simplifying it too much? because what I did seems to make sense but it looks like that's not what they wanted me to do.
 
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Isn't it just 0/2=0?
 


The original limit was neither of the indeterminate forms [0/0] or [infinity/infinity], so L'Hopital's Rule doesn't apply here. Your first approach was correct.
 

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