Proving that the pdf of reciprocal of random variable has zero density at the origin. 
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...
 A: 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 $Y.$
