Given a real-valued, continuous random variable $Y$ with an infinitely differentiable pdf $f_{Y}(y)$, define the reciprocal random variable $$ Z=\frac{1}{Y}. $$
Using the univariate transformation formula it can be shown that the pdf of $Z, f_{Z}$, is $$ f_{Z}(z) = \frac{f_{Y}(z^{-1})}{z^{2}}. $$
Under the above stated conditions, is $f_{Z}(0)=0$ generally true?
My Approatch
Intuitively, the only value $f_{Z}(0)$ could seemingly take on is $0$. So I attempted an initial start at a "proof" below...
$$ f_{Z}(0) = \lim_{z\to 0} f_{Z}(z) = \lim_{z\to 0} \frac{f_{Y}(z^{-1})}{z^{2}} $$
Let $t=1/z$, then
$$ f_{Z}(0) = \lim_{t\to\pm\infty}t^{2} f_{Y}(t) $$
Since $f_{Y}$ is a pdf, it must be a positive function with a finite integral meaning that $\lim_{t\to\pm\infty} f_{Y}(t)=0$. It can then be observed that, $$ \lim_{t\to\pm\infty}t^{2} f_{Y}(t)=\infty\times 0, $$ which is an indeterminate form. Therefore we rewrite the limit and apply L'Hospital's rule $$ f_{Z}(0) =\lim_{t\to\pm\infty}\frac{t^{2}}{f_{Y}^{-1}(t)} =\lim_{t\to\pm\infty}-\frac{2t}{\frac{f_{Y}'(t)}{f_Y^2(t)}} =\lim_{t\to\pm\infty}-\frac{2t f_Y^2(t)}{f_{Y}'(t)} $$
Not sure where to go from here...
Observe that $$ \int_{-\infty}^\infty \frac{dx}{1+x^2} = \pi $$ and suppose that for every measurable $A\subseteq\mathbb R$ $$ \Pr(Y\in A) = \int_A \frac{dx/\pi}{1+x^2}. \tag 1 $$ This is the standard Cauchy distribution. Let $Z= 1/Y.$ Then \begin{align} & f_Z(z) \, dz = d \Pr(Z\le z) = d \left. \begin{cases} \Pr( Y \le 1/z ) & \text{if } z<0, \\ \frac 1 2 + \Pr( Y\ge 1/z ) & \text{if } z>0, \end{cases} \right\} \\[10pt] = {} & d \left. \begin{cases} \displaystyle \int_{-\infty}^{1/z} \frac {dx/\pi}{1+x^2} & \text{if } z<0, \\[10pt] \displaystyle \frac 1 2 + \int_{1/z}^\infty \frac{dx/\pi}{1+x^2} & \text{if } z>0, \end{cases} \right\} = \frac{dz/\pi}{1+z^2}. \end{align} Thus the distribution of $Z=1/Y$ is the same as that of $Y,$ and the density at $0$ is not $0.$
Here is a geometric way of looking at it:
Beams of light shine north, south, east, and west from the point $(0,1).$ Then we make the object on which they are mounted rotate at a uniform rate. The proportion of the time that the initially north-south beams shine on the set $A$ on the $x$-axis is then given by the integral $(1)$ above. By symmetry, the same is true of the initially east-west beams. But the $x$-coordinate of the point on the $x$-axis illuminated by the initially east-west beam is $-1/Y$ when the point illuminated by the initially north-south beam is $Y.$