I was trying to solve this problem:
Show that, for large value of $x$, $\frac{1}{y} = \frac{2 + 3x^2}{3(1+x^2)^\frac{3}{2}} + \frac{1}{3} \approx 3 -\frac{9}{x}$. So, I looked that the mark scheme, it indicated that $(1+x^2)^\frac{3}{2} \approx (x^2)^\frac{3}{2}$ And $(2+3x^2) \approx 3x^2$. Then, we would get:
$\frac{1}{y} \approx \frac{1}{3} (1+\frac{3}{x})$.
So my question is, why does $(1+x^2)^\frac{3}{2} \approx (x^2)^\frac{3}{2}$ and $ (2+3x^2) \approx 3x^2$. Is there a general rule?
For example if we want to approximate $(a+bx^2)$ we would abandon a, and keep b. Is that so? If it is, may I know what is the mechanism behind this?
Thank you so much for you guys' replies.
I got $\frac{2 + 3x^2}{3(1+x^2)^\frac{3}{2}} + \frac{1}{3} \approx \frac{1}{3} + \frac{1}{x}$, actually.
But the larger point is this: What you are really using is $(1+x^2)^{\frac{3}{2}} = (1+\epsilon_1(x))(x^2)^{\frac{3}{2}}$ and $2+3x^2 = (1+\epsilon_2(x))(3x^2)$, where $\epsilon_1(x),\epsilon_2(x)$ are vanishing with $x$ large. So:
$$\frac{2 + 3x^2}{{3(1+x^2)^\frac{3}{2}}} = \frac{(1+\epsilon_2(x))3x^2}{3(1+\epsilon_1(x))(x^2)^{\frac{3}{2}}}$$
which as $x$ goes to infinity and $\epsilon_1(x)$ and $\epsilon_2(x)$ vanish, gives the result.
In fact, you could even put [say] $ax^2 + bx^4 = (1+\epsilon(x))bx^4$ where $\epsilon(x)$ goes to 0 as $x$ gets large, and on the other end, $ax^2+bx^4 = (1+\epsilon(x))ax^2$ where $\epsilon(x)$ goes to 0 as $|x|$ goes to 0.