How can the Taylor series help prove the limit of cosine?

[Damidami]

I have to prove that
\cos(x) = 1 - \frac{x^2}{2} + O(x^4) (x \to 0)

My ugly attempt:
\lim_{x \to 0} \frac{\cos(x) - 1 + \frac{x^2}{2}}{x^4}

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

\lim_{x \to 0} \frac{\sin(x)}{4x^3} + \frac{1}{2x^2}

\lim_{x \to 0} \frac{1}{4x^2} + \frac{1}{2x^2}

\lim_{x \to 0} \frac{1}{4x^2} + \frac{2}{4x^2} = \frac{3}{4x^2} = \infty (It should be a finite number)

Something does not work here, any help? Thanks!

 

[LCKurtz]

Have you looked at the Taylor series with remainder theorem?

 

[Damidami]

Have you looked at the Taylor series with remainder theorem?

Hi LCKurtz,
I was searching an alternative (and elementary) way to prove it (otherwise there is nothing to prove, right?)

My thought was: It should work using L'Hopital some finite numer of times, doesn't it? I don't know what am I doing wrong.

Thanks.

 

[LCKurtz]

Hi LCKurtz,
I was searching an alternative (and elementary) way to prove it (otherwise there is nothing to prove, right?)

My thought was: It should work using L'Hopital some finite numer of times, doesn't it? I don't know what am I doing wrong.

Thanks.

Yes, you can do it that way:

\lim_{x\rightarrow 0}\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}<br /> =\lim_{x\rightarrow 0}\frac{-\sin(x)+x}{4x^3}

and keep going as long as you have 0/0 form. You get 1/24 eventually.

 

[Damidami]

Yes, you can do it that way:

\lim_{x\rightarrow 0}\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}<br /> =\lim_{x\rightarrow 0}\frac{-\sin(x)+x}{4x^3}

and keep going as long as you have 0/0 form. You get 1/24 eventually.

You are completely right! Thanks!

I think my mistake was in the second step:

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

The limit of a sum is the sum of the limit only of both limits exists right? But the second one clearly is +\infty and that isn't a real number, so that theorem does not apply, am I right now?

Thanks!