Help Finding Limit: Homework Statement

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SUMMARY

The limit of the function as x approaches 0, specifically \lim_{x \to 0} (1 - \cos x), leads to an undefined expression when directly substituting x=0, as it results in ln(1-1). However, through analytical methods, particularly L'Hôpital's Rule, it can be shown that \lim_{x \to 0^{+}} \ln(1 - \cos x) = -\infty. This indicates that the limit does not exist unless both left and right limits agree, which they do in this case, confirming the result.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with the cosine function and its behavior near zero
  • Basic knowledge of natural logarithms and their properties
  • Introduction to L'Hôpital's Rule for evaluating indeterminate forms
NEXT STEPS
  • Study the application of L'Hôpital's Rule in detail
  • Explore the behavior of trigonometric functions near their limits
  • Learn about the properties of logarithmic functions
  • Practice solving limits involving trigonometric and logarithmic expressions
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Students beginning their calculus journey, particularly those struggling with limits and the application of L'Hôpital's Rule. This discussion is beneficial for anyone looking to solidify their understanding of limits involving trigonometric and logarithmic functions.

fishingspree2
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Homework Statement


\mathop {\lim }\limits_{x \to 0} (1 - \cos x)

2. The attempt at a solution

well as x goes to 0, cos x goes to 1... ln(1-1) is undefined
now if I forget about plugging in x=0 and think a little bit, the ln argument gets very very small, and the logarithm of a decimal number is a negative number... so I would say the limit is minus infinity.

however, is it possible to get that result analytically... by transforming/simplyfing/etc. the function?

sorry, I have just finished precalculus and I am beginning calculus, so my calculus skills are crap =\

thank you
 
Last edited:
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hmm ... problem is: \lim_{x\rightarrow0}(1-\cos x)?

Can't you just plug it in?

Or is it:

\lim_{x\rightarrow0}\ln{(1-\cos x)}

Ok so have the graph infront of you. You know as you "approach" coming from the right, it's negative infinity ...

\lim_{x\rightarrow0^{+}}\ln{(1-\cos x)}=-\infty

What about from the left? Because in order for the Limit to exist, the L & R Limits must agree. But for the natural logarithm, how is it uniquely defined?
 
Last edited:
Once you have learned L^Hopitals rule, you will be able to apply it to this equation and get the same answer analytically.
 

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