Ratio Limit involving $t-$ distribution If $X$ is a random variable with t- distribution of parameter $\mathcal{v}$, how can I prove that
$$
\lim_{\alpha \to 1^{+}} \frac{ES_{\alpha}(X)}{VaR_{\alpha}(X)} = \frac{\mathcal{v}}{\mathcal{v}-1}
$$
Where accordingto the book  Quantitative Risk Management: Concepts, Techniques and Tools by  Alexander J. McNeil  et al, the formulas are given by: 
$$ES_{\alpha} = \frac {g_v(t_v^{-1}(\alpha))}{1- \alpha} \frac{v+(t_v^{-1}(\alpha))^2}{v-1}$$
 and 
$$VaR_{\alpha} = t_v^{-1}(\alpha)$$
where $t_v$ denotes the distribution function of the standard student and $g_v$ the density function of the same distribution
 A: I will first transform the argument of the limit, since this will help in solving the problem. I introduce a new variable $x=t_v^{-1}(\alpha)$ such that we obtain
$$\lim_{\alpha \to 1^{+}} \frac{ES_{\alpha}(X)}{VaR_{\alpha}(X)} = \lim_{\alpha \to 1^{+}} \frac{g_v(t_v^{-1}(\alpha))}{(1- \alpha)t_v^{-1}(\alpha)} \frac{v+(t_v^{-1}(\alpha))^2}{v-1} = \lim_{x \to +\infty} \frac{g_v(x)}{(1- t_v(x))x}\frac{v+x^2}{v-1}$$
I pull out the factor $v/(v-1)$ from the last limit to get
$$\lim_{\alpha \to 1^{+}} \frac{ES_{\alpha}(X)}{VaR_{\alpha}(X)} = \frac{v}{v-1}\lim_{x \to +\infty} \frac{g_v(x)\left(1+\frac{x^2}{v}\right)}{(1- t_v(x))x}$$
So, we have to prove that the latter limit is 1 and we're done. I slightly rewrite the limit
$$\lim_{x \to +\infty} \frac{\frac{1}{x}g_v(x)\left(1+\frac{x^2}{v}\right)}{1- t_v(x)} = \lim_{x \to +\infty} \frac{\frac{1}{x}c_v\left(1+\frac{x^2}{v}\right)^{-\frac{v-1}{2}}}{1- c_v\int_{-\infty}^x\left(1+\frac{u^2}{v}\right)^{-\frac{v+1}{2}}du}$$
Using the explicit form of the density $g_v$ of the student-t distribution. I just don't make the normalization constants explicit, since they will play no role in the rest of the computation. I denoted them by $c_v$. Now we apply l'Hôpitals rule, obtaining
$$\lim_{x \to +\infty} \frac{-\frac{1}{x^2}c_v\left(1+\frac{x^2}{v}\right)^{-\frac{v-1}{2}}+\frac{1}{x}c_v\left(-\frac{v-1}{2}\right)\left(1+\frac{x^2}{v}\right)^{-\frac{v+1}{2}}\frac{2x}{v}}{- c_v\left(1+\frac{x^2}{v}\right)^{-\frac{v+1}{2}}}$$
You see that indeed the factor $c_v$ cancels out, and we can also put the denominator in the numerator to get
$$\lim_{x \to +\infty} \frac{1}{x^2}\left(1+\frac{x^2}{v}\right)+\frac{1}{x}\left(\frac{v-1}{2}\right)\frac{2x}{v} = \lim_{x \to +\infty}1+\frac{1}{x^2}=1$$
And we're done.
