Using Taylor expansion for limit solving

[Hernaner28]

Hi. I just want to ask: how can I realize that I need to do the 4th order taylor's expansion for solving a precise limit? e.g.

\mathop {\lim }\limits_{x\to 0} \frac{{{e^x}-1-\frac{{{x^2}}}{2}+\sin x-2x}}{{1-\cos x-\frac{{{x^2}}}{2}}}

We need the 4th order of the expansion but how can I realize just seeing it?another question. Why is this true:
\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{o(x)}}{x} = 0?

o(x) is the remainder (Landau notation)
Thanks!

 

[Office_Shredder]

Your tex isn't working for me so I can't answer your first question, but in general you need to use enough terms so that all larger terms just become zero in the limit. Try re doing the calculation using a 5th degree taylor polynomial, our a sixth degree one and you should get the same answer. With time it will become intuitive how many terms you will need fire this to happen

 

[Hernaner28]

Hmm, alright and what about the remainder o(x)?

PD: I've fixed the LaTeX just when you were typing I guessThank you1

 

[vela]

Why is this true:
\lim_{x \to 0} \frac{{o(x)}}{x} = 0?o(x) is the remainder (Landau notation)

Remainder of what?

 

[Hernaner28]

Of the Taylor polynomial

 

[Hernaner28]

Tried to solve the following but couldn't:

\begin{array}{c}<br /> \mathop {\lim }\limits_{x\to 0} \frac{{\log (1+x)-x-\frac{{{x^2}}}{2}}}{{\tan x-\sin x}}=\frac{{-2x-{x^2}-\frac{{{x^3}}}{3}+{o_1}({x^3})}}{{2x+\frac{{{x^3}}}{3}+{o_2}({x^3})+{o_3}({x^3})}}\\<br /> =\frac{{\frac{{-2x-{x^2}-\frac{{{x^3}}}{3}+{o_1}({x^3})}}{{{x^3}}}}}{{\frac{{2x+\frac{{{x^3}}}{3}+{o_2}({x^3})+{o_3}({x^3})}}{{{x^3}}}}}\\<br /> =\frac{{\frac{{-2}}{{{x^2}}}-\frac{1}{x}-\frac{1}{3}+\overbrace {\frac{{{o_1}({x^3})}}{{{x^3}}}}^0}}{{\frac{2}{{{x^2}}}+\frac{1}{6}+\underbrace {\frac{{{o_2}({x^3})}}{{{x^3}}}}_0+\underbrace {\frac{{{o_3}({x^3})}}{{{x^3}}}}_0}}\\<br /> =?<br /> \end{array}

EDIT: I could solve it. It's -infinity. Why is this true:
\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{o(x)}}{x} = 0?

o(x) is the remainder (Landau notation)
Thanks!

 

[vela]

It's true by definition of Landau notation, but that's probably not the answer you're looking for. When you write, say, o(x), you're talking about the terms that go to 0 faster than x does. In the quotient o(x)/x, there are two competing behaviors. The numerator going to 0 tends to drive the quotient toward 0, but the denominator going to 0 tends to cause the quotient to blow up. The behavior that dominates determines what happens. In this case, o(x), by definition, goes to 0 faster than x does, so o(x)/x goes to 0 in the limit that x goes to 0.

Consider the limit
$$\lim_{x \to 0} \frac{e^x-1}{x} = \lim_{x \to 0} \frac{\left(1+x+\frac{x^2}{2!}+\cdots\right) - 1}{x} = \lim_{x \to 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x} = \lim_{x \to 0} \frac{x + o(x)}{x}$$ The quadratic and higher-order terms go to 0 faster than x does, which is why you can replace them with o(x). If you skip the o-notation, though, you can see what happens is simply this:
$$\lim_{x \to 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x} = \lim_{x \to 0} \frac{x + x\left(\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)}{x} = \lim_{x \to 0} \left[1 + \left(\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)\right]$$ All the terms that correspond to o(x)/x, the stuff inside the parentheses in the last limit, have a factor of x in them, so as x approaches 0, they vanish.

 

[Hernaner28]

But I'm not replacing the Taylor expansion by o(x) as you have just done. We prooved that the remainder of order n is $$o((x-a)^n))$$ when x->a. See how I solved the limit above your post. I just added the remainder to the Taylr polynomial (o(x))

Could it be that, as the remainder tends to be zero, then the quotient o(x)/x is zero?

 

[vela]

I didn't replace the Taylor expansion by o(x).