# Using Taylor expansion for limit solving

1. Apr 26, 2012

### 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!

Last edited: Apr 26, 2012
2. Apr 26, 2012

### Office_Shredder

Staff Emeritus
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

3. Apr 26, 2012

### Hernaner28

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

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

Thank you1

4. Apr 26, 2012

### vela

Staff Emeritus
Remainder of what?

5. Apr 26, 2012

### Hernaner28

Of the Taylor polynomial

6. Apr 26, 2012

### Hernaner28

Tried to solve the following but couldn't:

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

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!

Last edited: Apr 26, 2012
7. Apr 26, 2012

### vela

Staff Emeritus
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.

Last edited: Apr 26, 2012
8. Apr 26, 2012

### 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?

9. Apr 26, 2012

### vela

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