Finding a limit without L'Hospital's Rule

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In summary, the limit of the given expression is solved by using the Taylor expansion for cosine. However, if the Taylor expansion is not known, the problem can be solved by writing the numerator as a sum and multiplying the numerator and denominator by the conjugate of the sum. Simplification can then be done to find the solution.
  • #1
mr.tea
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Homework Statement


Hi,
I have been trying to find the limit of the following expression in order to determine the Big-o(in particular, the order of decaying to zero)

[tex]\lim_{x\rightarrow 0} \frac {\cos (x) -1 +\frac {x}{2} }{x^4}[/tex]

Is there another "reasonable" way to solve it?

Thank you,
Thomas

Homework Equations


[tex]\lim_{x\rightarrow 0} \frac {\cos (x) -1 +\frac {x}{2} }{x^4}[/tex]

The Attempt at a Solution


I have tried to factorize x^2 in the numerator and denominator, and yield nothing(well, yield that the limit goes to infinity,). tried to replace cosx with sqrt(1-sin^2 x) and multiply with the conjugate but nothing.
 
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  • #2
This limit doesn't exist, as can be seen from the Maclaurin series of ##\cos##. Maybe you wanted ##\frac{x^2}{2}## instead of ##\frac{x}{2}##?
 
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  • #3
What's a simple equivalent in 0 of ##\cos(x)## ?
 
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  • #4
mr.tea said:

Homework Statement


Hi,
I have been trying to find the limit of the following expression in order to determine the Big-o(in particular, the order of decaying to zero)
[tex]\lim_{x\rightarrow 0} \frac {\cos (x) -1 +\frac {x}{2} }{x^4}[/tex]Is there another "reasonable" way to solve it?

Thank you,
Thomas

Homework Equations

[tex]\lim_{x\rightarrow 0} \frac {\cos (x) -1 +\frac {x}{2} }{x^4}[/tex]

The Attempt at a Solution


I have tried to factorize x^2 in the numerator and denominator, and yield nothing(well, yield that the limit goes to infinity,). tried to replace cosx with sqrt(1-sin^2 x) and multiply with the conjugate but nothing.
It would be a much more interesting problem if that was x2/2 in the numerator.

Are you sure that shouldn't be :
##\displaystyle \ \lim_{x\to 0} \frac {\cos (x) -1 +\displaystyle \frac {x^2}{2} }{x^4} \ ##​
 
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  • #5
Krylov said:
This limit doesn't exist, as can be seen from the Maclaurin series of ##\cos##. Maybe you wanted ##\frac{x^2}{2}## instead of ##\frac{x}{2}##?

SammyS said:
It would be a much more interesting problem if that was x2/2 in the numerator.

Are you sure that shouldn't be :
##\displaystyle \ \lim_{x\to 0} \frac {\cos (x) -1 +\displaystyle \frac {x^2}{2} }{x^4} \ ##​

oh! sorry! it is like that!
it is so embarrassing!

The Equation is with x^2/2...

Sorry about that.

Thomas
 
  • #6
mr.tea said:
oh! sorry! it is like that!
it is so embarrassing!

The Equation is with x^2/2...

Sorry about that.

Thomas
The quickest way to get a solution is to use the Taylor expansion for cosine.
 
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  • #7
SammyS said:
The quickest way to get a solution is to use the Taylor expansion for cosine.

Thank you for your answer.

We have not learned Taylor expansion yet. So, is there a reasonable way to solve it with simple algebraic steps?

(well, this question is not so important if there is no reasonable way to solve it, so don't feel obliged to give a full solution. Idea will be enough. I have solved it already by L'H., but I am probably a masochist :headbang: )
 
  • #8
mr.tea said:
Thank you for your answer.

We have not learned Taylor expansion yet. So, is there a reasonable way to solve it with simple algebraic steps?

(well, this question is not so important if there are no reasonable way to solve it . I have solved it already by L'H., but I am probably a masochist :headbang: )

Write the numerator as ##\displaystyle \ \cos (x) +\left(\displaystyle \frac {x^2}{2} -1\right) \ ##.

Then multiply the numerator & denominator by ##\displaystyle \ \cos (x) -\left(\displaystyle \frac {x^2}{2} -1\right) \,,\ ## to get a difference of squares in the numerator.

Do some simplifying & see what pops out.
 

Related to Finding a limit without L'Hospital's Rule

1. What is the concept of finding a limit without using L'Hospital's Rule?

Finding a limit without using L'Hospital's Rule involves using other methods, such as algebraic manipulation or substitution, to evaluate the limit of a function at a specific point. This is useful when L'Hospital's Rule cannot be applied, such as when the limit involves a product or quotient of two functions.

2. How do I know when I can't use L'Hospital's Rule to find a limit?

L'Hospital's Rule can only be used when the limit involves a fraction where the numerator and denominator both approach zero or infinity. If the limit does not fit this form, then L'Hospital's Rule cannot be applied and alternative methods must be used.

3. Can I always find a limit without using L'Hospital's Rule?

Yes, there are many methods for finding limits without using L'Hospital's Rule. It may require more algebraic manipulation or creativity, but a limit can always be evaluated without using L'Hospital's Rule.

4. Is it necessary to find a limit without using L'Hospital's Rule?

No, L'Hospital's Rule is a useful tool for evaluating limits, but it is not always necessary. In some cases, it may be simpler and more efficient to use L'Hospital's Rule, while in others, alternative methods may be more appropriate.

5. Are there any disadvantages to not using L'Hospital's Rule to find a limit?

One disadvantage of not using L'Hospital's Rule is that it may require more time and effort to evaluate the limit using alternative methods. Additionally, L'Hospital's Rule may provide a more accurate approximation of the limit in some cases. However, it is important to understand and be able to use other methods for finding limits without relying solely on L'Hospital's Rule.

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