Consider the following function
$$f_{\alpha}(\rho)=\frac{1}{(1+\alpha)^3-1}\Big(1+\frac{1}{2}\frac{(1+\alpha)^3}{\rho^3}\Big)\tag{1}$$
where $\alpha\in\mathbb{R}^+$ is a parameter and $\rho$ is an independent variable which lies in $1\le\rho\le1+\alpha$. This function appears for the radial stress in a thick walled spherical shell and $\alpha$ represents the dimensionless thickness of the shell and $\rho$ is the dimensionless radial coordinate. For more details you can take a look at this post on Physics SE.
My goal is to make an approximation of this function when $\alpha$ becomes too small, that is when the shell is too thin. If I take the limit $\alpha\to0^+$ then $\rho\to 1^+$ and $f_{\alpha}(\rho)\to+\infty$ so I get nowhere with this. I re-write this as follows
$$f_{\alpha}(\rho)=\frac{1}{\alpha}\frac{\alpha}{(1+\alpha)^3-1}\Big(1+\frac{1}{2}\frac{(1+\alpha)^3}{\rho^3}\Big)\tag{2}$$
and I decide to take the limit of the coefficient of $\frac{1}{\alpha}$ in $(2)$ which leads to
$$f_{\alpha}(\rho)\approx \bar f_{\alpha}(\rho):=\frac{1}{\alpha}\lim_{\alpha\to 0^+}\alpha f_{\alpha}(\rho)=\frac{1}{2\alpha}\tag{3}$$
This is the answer which is used in engineering text books.
Does my approach make sense? Is there a better way to make an approximation?
I think if we show that
$$\lim_{\alpha\to 0^+}\Big(f_{\alpha}(\rho)-\bar f_{\alpha}(\rho)\Big)=0 \tag{4}$$
then every thing seems fine.
Probably off-topic but too long for a comment.
If we consider $$f=\frac{1}{(1+\alpha)^3-1}\Big(1+\frac{1}{2}\frac{(1+\alpha)^3}{\rho^3}\Big)$$ where $\rho$ is a constant, a simple Taylor series built at $\alpha=0$ would give $$f=\frac{\frac{1}{\rho ^3}+2}{6 \alpha }+\frac{1}{3} \left(\frac{1}{\rho ^3}-1\right)+\frac{1}{9} \left(\frac{1}{\rho ^3}+2\right)\alpha+O\left(\alpha ^2\right)$$ For sure, the problem becomes more complex if $\rho$ depends on $\alpha$.
Taking into account the bounds, let us define $\rho=1+\phi(\alpha)$ with $\phi(0)=0$. This would lead to $$f=\frac{1}{2 \alpha }-\frac{1}{2}\phi '(0)+ \left(\frac{1}{3}+\phi '(0)^2-\phi '(0)-\frac{1}{4}\phi ''(0)\right)\alpha+O\left(\alpha ^2\right)$$
Now, making series of series and using $\rho=x+1$, this would give $$f=\frac{x^2-\frac{x}{2}+\frac{1}{2}}{\alpha }+\left(2 x^2-x\right)+\frac{1}{3} \left(2 x^2-x+1\right)\alpha+O\left(\alpha ^2\right)$$