I got a function $f(x) = \frac {k}{(a^2 + x^2)^{ {3}/{2}}}\vec e_x$, where $\vec e_x$ is an unit vector in the $x$ direction and $k$ is a constant. And I should study the asymptotic behaviour of this function for $x \to +\infty$ and $x \to -\infty$.
I tried to graph this function and I didn't find any critic points and I calculated this limit as $0$. So I don't understand what to do, If I missed something important.
The function corresponds to the magnetic induction and this example is from an old test, so I don't have solution and I am curious.
I have searched that asymptotic analysis and do I understand right that I need to find a function that this function relates to for really large x? This part of math/physics is new to me, so that is why I was confused that the limit is just $0$ and I did not find a other counterexample to understand that.
Using the Taylor series for $(1+x)^{-3/2}$, we get $$ \begin{align} \frac{k}{\left(a^2+x^2\right)^{3/2}} &=k|x|^{-3}\left(1+\frac{a^2}{x^2}\right)^{-3/2}\\ &=k|x|^{-3}-\frac{3ka^2}2|x|^{-5}+\frac{15ka^4}8|x|^{-7}+O\!\left(|x|^{-9}\right) \end{align} $$ So asymptotically, to the first order, we have $$ \frac{k}{\left(a^2+x^2\right)^{3/2}}\sim k|x|^{-3} $$ This means that $$ \lim_{|x|\to\infty}\frac{\frac{k}{\left(a^2+x^2\right)^{3/2}}}{k|x|^{-3}}=1 $$ For higher order of approximation, we have the asymptotic series $$ \frac{k}{\left(a^2+x^2\right)^{3/2}}\sim\frac{k}{|x|^3}\sum_{j=0}^{n-1}(2j+1)\binom{2j}{j}\left(-\frac{a^2}{4x^2}\right)^j $$ Unlike most asymptotic series, the series above actually converges for $|x|\gt|a|$.