Finding a limit without L'Hospital's Rule

[mr.tea]

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)

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

Is there another "reasonable" way to solve it?

Thank you,
Thomas

Homework Equations

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

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.

 

[S.G. Janssens]

This limit doesn't exist, as can be seen from the Maclaurin series of $\cos$. Maybe you wanted $\frac{x^2}{2}$ instead of $\frac{x}{2}$?

 

[geoffrey159]

What's a simple equivalent in 0 of $\cos(x)$ ?

 

[SammyS]

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)
\lim_{x\rightarrow 0} \frac {\cos (x) -1 +\frac {x}{2} }{x^4}Is there another "reasonable" way to solve it?

Thank you,
Thomas

Homework Equations

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

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} \$​

 

[mr.tea]

This limit doesn't exist, as can be seen from the Maclaurin series of $\cos$. Maybe you wanted $\frac{x^2}{2}$ instead of $\frac{x}{2}$?

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

 

[SammyS]

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.

 

[mr.tea]

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 )

 

[SammyS]

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 )

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.