Question: Let $Y= 1-X$. Find its cumulative distribution function and its density, in terms of $F_X$ and $f_X$
My way to do:
$$F_{Y}(y)=P(1-X<y)=P(X>1-y)\\=1-P(X\le1-y)\\=1-F_X(1-y) \\=1-\lim \limits_{x \to 1-y} F_X(x)$$
$$F_{Y}(y)=1-\lim \limits_{x \to 1-y} F_X(x)$$
Then, I want to find PDF by $\frac{d(F_{Y}(y))}{dy}$.
Is PDF just $-\lim \limits_{x \to 1-y}f_X(x)$ ? If wrong, how can I get that.
On
You have already obtained $$F_{Y}(y)=1-F_X(1-y)$$
(However, I am not sure if it is necessary to express the same result using the limit concept)
Here $Y= 1-X$
$\Rightarrow X=1-Y$
So, $\frac{dx}{dy}=-1$
Now, you can differentiate the c.d.f. of $Y$ with respect to $y$ which yields
$$\frac{d}{dy}F_{Y}(y)=\frac{d}{dx}[1-F_X(1-y)]\frac{dx}{dy}$$
$$\Rightarrow f_Y(y)=[-f_X(1-y)](-1)$$
$$\Rightarrow f_Y(y)=f_X(1-y)$$
$$F_Y(y)=P(1-X\leq y)=P(X\geq1-y)=1-P(X<1-y)=1-P(X\leq1-y)=$$$$1-F_X(1-y)\tag1$$where the $4$-th equality is implied by the fact that $F_X$ is continuous.
This is enough concerning CDF: $F_Y$ is expressed by means of $F_X$.
If $F_X(x)$ would be differentiable (we don't know) having derivative $f(x)$ then $(1)$ tells us that $F_Y(y)$ is differentiable having derivative $f(1-y)$.
This makes us "suspect" that the function $g(y)=f_X(1-y)$ will serve as PDF of $Y$.
This can be verified by:$$\int_{-\infty}^yf_X(1-z)dz=\int_{1-y}^{\infty}f_X(x)dx=P(X\geq1-y)=P(Y\leq y)$$
Proved is now that: $$f_Y(y)=f_X(1-y)$$
This is enough concerning PDF: $f_Y$ is expressed by means of $f_X$.