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I have one maths challenge which I could not resolve. The question is the following. Let f(x) denotes the probability density of a continuous RV., X. I want to know the limit f(x) as x approaches to either -∞ or +∞. It might seem silly, but I have do idea how to do that.

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    @Riccardo.Alestra: I would like to know the case of a generalised distribution. But if that isn't possible, I will go for the normal, exponential and log-normal distributions2012-07-06

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It is not necessary that a density have any limit as $x\to\infty$ or $x\to -\infty$. The sole criteria for a function $f$ to be a density are $f\ge 0$ a.e. and that $\int_{-\infty}^\infty f(x)\mbox{d}x = 1.$

Such a function could have high, narrow spikes at either or both ends. It is, in fact, possible to have $\limsup_{n\to\infty} f(x) = \infty$ and $\limsup_{n\to-\infty} f(x) = \infty.$ Try to construct such an example. It is not terribly hard.

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    what about the continuity of $f$?2014-04-06
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Assuming that this question arose in an undergraduate (non-measure-theoretic) course in probability, the answer that the OP (or perhaps the OP's instructor) might have been looking for might be that $\lim_{x\to\infty} f(x) = \lim_{x\to -\infty} f(x) = 0.$ The rationale would be that if $f(x)$ were converging to a positive number $c$ as $x \to \infty$, then $\int_{-\infty}^{\infty}f(x)\,\mathrm dx$ would diverge. In more detail for the OP's benefit, there would exist an $x_0$ such that for all $x > x_0$, $|f(x) - c| < \frac{c}{2} \Rightarrow f(x) > \frac{c}{2}$, and so $\int_{-\infty}^{\infty}f(x)\,\mathrm dx \geq \int_{x_0}^{\infty}f(x)\,\mathrm dx \geq \int_{x_0}^{\infty}\frac{c}{2}\,\mathrm dx.$ Indeed, not only must $f(x)$ converge to $0$ as $x \to \pm\infty$ but the convergence must be faster than $|x|^{-1}$.

Of course, since $f(x)$ can be viewed as a representative member of a class of density functions that differ only on a set of measure zero, one can find other functions in this class for which $\limsup f(x) = \infty$, e.g. $fx) = \begin{cases} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right), &x ~\text{is irrational},\\ |x|, &x ~\text{is rational.} \end{cases}$