Solving the Limit of cosx-1/x: Tips & Tricks

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In summary, the conversation discusses finding the value of the limit x-->0 cosx-1/x. The original poster is unsure of how to solve it and asks for tips. Another user suggests using L'Hospital's Rule or the cosine series expansion. The original poster then uses trigonometric identities to manipulate the expression and eventually concludes that the limit is equal to 0. The other user confirms that this is correct and the conversation ends with gratitude from the original poster.
  • #1
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Homework Statement



Find the value of
limit x-->0 cosx-1/x


The Attempt at a Solution


one thing i understood is that i can't use the triangle and circle way as i did for
limit x-->0 sinx/x

nor have i found out a way to use the 'sandwich' theorem.
So anyone have any tips please?

I have a guess for it to be 0, but yet, i don't know.
 
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  • #2
Is it (cosx-1)/x? Then your guess is correct.
 
  • #3
yea, but how do i solve it then?
 
  • #4
Ever heard of L'Hospital's Rule?
 
  • #5
nope
 
  • #6
Cosine series?I mean expansion of cosx as x-->0?
 
  • #7
umm...
if x-->0
then,
cosx has limit 1, right?(sorry, but i don't know much calculus.almost none..if you could explain a bit,please )
 
  • #8
Alright i will avoid unnecessary calculus. How much trigonometry do you know? Do you know
cos2x = cos2x-sin2x = 2cos2x-1 = 1-2sin2x
Using these formulae a little manipulation and your stated identity sinx/x;x-->0 = 1 try to work out a solution. i will be around if you need help.
Hint:Write cosx as cos2(x/2).
 
  • #9
yeah i know that trig
so...
using that hint it becomes

=-2sin^2(x/2)/x

= - sin^2(x/2)/(x/2)

= - sin(x/2) * sin(x/2)/(x/2)

but since limit of sin(x/2) as x tends to 0 is 0...hence the whole is 0.

Is this right?
 
  • #10
Right on the mark.
 
  • #11
whew!
Thanks a lot, clever Orange! :smile:
 

FAQ: Solving the Limit of cosx-1/x: Tips & Tricks

What is the limit of cosx-1/x as x approaches 0?

The limit of cosx-1/x as x approaches 0 is undefined. This is because when x approaches 0, the denominator becomes 0 which results in division by 0, which is undefined.

Can L'Hopital's rule be used to solve the limit of cosx-1/x?

Yes, L'Hopital's rule can be used to solve the limit of cosx-1/x. This rule is used when the limit results in an indeterminate form, such as 0/0 or ∞/∞. In this case, the limit of cosx-1/x as x approaches 0 is in the form of 0/0, so L'Hopital's rule can be applied.

What are some tips for solving the limit of cosx-1/x?

One tip for solving this limit is to use the trigonometric identity cosx-1 = -2sin²(x/2). This allows you to rewrite the limit as -2sin²(x/2)/x which can then be simplified using L'Hopital's rule. Another tip is to use a graphing calculator to visualize the function and better understand its behavior near the limit point.

Are there any other methods besides L'Hopital's rule for solving this limit?

Yes, there are other methods for solving this limit. One method is to use the Taylor series expansion of cosx and then evaluate the limit using the properties of limits. Another method is to use the squeeze theorem by finding two other functions that are always between cosx-1/x and its limit as x approaches 0.

Why is it important to understand how to solve the limit of cosx-1/x?

Understanding how to solve this limit is important because it is a common problem in calculus and can also be applied in various real-world scenarios. It also helps in understanding the behavior of functions near a certain point and can be a useful tool in solving more complex problems involving limits. Additionally, mastering this limit can improve overall understanding of calculus and its applications in other fields of science.

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