Would anyone be kind to explain to me how to do this types of problems?
My proffessor gave us couple of this problems for homework and we dont have solutions. This is my first time to do something like this. This is the easiest of them. Please write me a solution and I will find pattern myself.
Thanks in advance.
On
Well, you can get $a$ from the fact that $f(x)/x\to a$.
After, you can use the fact that $f(x)-ax \to b$ to get $b$.
Can you get $c$ using the same method?
On
You can boil it down to the binomial series as follows:
$$f(x)\stackrel{x>0}{=}(1+x)(1+\frac 1x)^{\frac 12}$$
Now, expand the binomial till order $2$:
$$f(x) = (1+x)\left(1+\frac 1{2x} - \frac 1{8x^2} + o\left(\frac 1{x^2}\right)\right)$$
Now, multiply and collect
$$f(x) = x+\frac 32 + \frac 3{8x} + o\left(\frac 1{x}\right)$$
For $x>-1$, we can write
$$\sqrt{\frac{(x+1)^3}{x}}=(x+1)\sqrt{1+\frac1x}\tag1$$
As $x\to \infty$, $t=1/x \to 0$. So, let us examine the behavior of the function $f(t)=\sqrt{1+t}$ for "small" values of $t$.
Let $e(t)$ denote the function $e(t)=\sqrt{1+t}-a-bt-ct^2$ so that $\sqrt{1+t}=a+bt+ct^2+e(t)$.
Note that $e(t)$ is the error between $\sqrt{1+t}$ and a quadratic polynomial $a+bt+ct^2$ that we view as an "approximation" for $\sqrt{1+t}$. Hence, we have
$$e(t)=\frac{(1-a^2)+(1-2ab)t+(2ac+b^2)t^2-2bct^3-c^2t^4}{\sqrt{1+t}+a+bt+ct^2}\tag2$$ Now for $a=1$, $b=1/2$, and $c=1/8$, $(2)$ becomes
$$e(t)=-\frac1{64}\left(\frac{t-8}{\sqrt{1+t}+1+t/2-t^2/8}\right)t^3$$ and the error function behaves like $Ct^3$ as $t\to 0$ (i.e., $e(t)=O(t^3)$). To see this, simply observe that $\lim_{t\to 0}e(t)/t^3=\frac1{16}$ and so, $e(t)=O(t^3)$ indeed. Moreover, we find that
$$\sqrt{1+t}=1+\frac12t-\frac18 t^2+O(t^3)\tag3$$
Finally, letting $x=\frac1t$ in $(3)$ and substituting in $(1)$ reveals
$$\sqrt{\frac{(x+1)^3}{x}}=(1+x)\left(1+\frac1{2 x}-\frac1{8x^2}+O\left(\frac1{x^3}\right)\right)$$
And now you can finish.