Finding Polynomials of Lowest Degree that is similar to $f$ as $x\to a$

Question

I'm working through the book "Asymptotic Analysis and Perturbation Theory" by William Paulsen and am having some trouble with the following problems.

Definition 1.1 p.1. Given two functions, $f(x)$ and $g(x),$ we say that $f(x)$ is similar to $g(x)$ as $x$ approoaches $a,$ written $$f(x)\sim g(x)\quad\text{as}\quad x\to a,$$ if $$\lim_{x\to a}\frac{f(x)}{g(x)}=1.$$

Problems 13, 15, 17, 19 p.8. Find the polynomial $p(x)$ of lowest degree that is similar to the following functions as $x\to a$.

1. $x^2,\ a=2$
2. $\cos x,\ a=\pi /2$
3. $2x^3-3x^2+1,\ a=1$
4. $1+\cos x,\ a=\pi$

1. I thought the answer would be $p(x)=4(x-1)$ but the book tells me it's $4$.

2. The line tangent to the curve at $x=\pi/2$ is $p(x)=\pi/2-x$, which agrees with the book.

3. The book tells me $3(x-1)^2$ but I do not know how they ended up there.

4. The answer is $(x-\pi)^2/2$, but this too I'm not quite sure how they obtained.

Can someone explain? I think I need help finding the order required for $p(x)$.

Answer 1

This polynomial is the first non-zero monomial in $h=x-a$ in the Taylor expansion as $x\to a,$ i.e. $h\to0.$

1. $p(x)=a^2=4.$
2. $\cos(\pi/2+h)=\sin(-h)\sim-h$ hence $p(x)=-h=\pi/2-x.$
3. $2(1+h)^3-3(1+h)^2+1=0+0h+3h^2+2h^3$ hence $p(x)=3h^2=3(x-1)^2.$
4. $1+\cos(\pi+h)=1-\cos h\sim h^2/2$ hence $p(x)=h^2/2=(x-\pi)^2/2.$