limits of continuous probability density function

Question

hello i have to find a limit of a probability distribution function. the FX(x) are not informed in the assignment, as how large or anything else.

and there are noting in my textbook about it.

my 2 problems are

$\lim\limits_{x \to -\infty} \frac{F_X(x)^2-1}{F_X(x)+1}$ negative infinity

and

$\lim\limits_{x \to \infty} \frac{F_X(x)^2-1}{F_X(x)+1}$ positive infinity

my solution are there are rules for limits of a density function as $\lim\limits_{x \to -\infty}=0$ and $\lim\limits_{x \to \infty}=1$

so would a solution not be

$\lim\limits_{x \to -\infty} \frac{F_X(x)^2-1}{F_X(x)+1}=\frac{0^2-1}{0+1}=-1$

and

$\lim\limits_{x \to \infty} \frac{F_X(x)^2-1}{F_X(x)+1}=\frac{1^2-1}{1+1}=\frac{0}{2}=0$

Answer 1

Note that for the $-\infty$ you could expand the numerator:$ (F-1)(F+1)$ and then cancel the denominator.

Answer 2

Your answer seems consistent with

• $\frac{a^2-1}{a+1}=a-1$ if $a\not=-1$. In particular this is $-1$ when $a=0$ and $0$ when $a=1$

• for a cumulative distribution function (not a density): $\lim\limits_{x \to -\infty} F_X(x) = 0$ and $\lim\limits_{x \to +\infty} F_X(x) = 1$

Answer 3

Notice that $$ \frac{F_X(x)^2-1}{F_X(x)+1} = \frac{(F_X(x)-1)(F_X(x)+1)}{F_X(x)+1}=F_X(x)-1 $$