# If M <mrow class="MJX-TeXAtom-ORD"> Z <mrow class="MJX-TeXAtom-O

If
${M}_{{Z}_{n}}\left(t\right)={\left(\frac{3{e}^{\frac{-t}{4\sqrt{n}}}+{e}^{\frac{3t}{4\sqrt{n}}}}{4}\right)}^{2n}$
Calculate $\underset{n\to \mathrm{\infty }}{lim}{M}_{{Z}_{n}}\left(t\right)$ and interpret the result
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jmibanezla
This has the ${1}^{\mathrm{\infty }}$ form.
So you can use this
$\underset{x\to a}{lim}f\left(x{\right)}^{g\left(x\right)}=\mathrm{exp}\left(\underset{x\to a}{lim}\left(f\left(x\right)-1\right)g\left(x\right)\right)$
when $\underset{x\to a}{lim}f\left(x{\right)}^{g\left(x\right)}$ is ${1}^{\mathrm{\infty }}$ form.
Using the above we have :
$\mathrm{exp}\left(\underset{n\to \mathrm{\infty }}{lim}\frac{3{e}^{\frac{-t}{4\sqrt{n}}}+{e}^{\frac{3t}{4\sqrt{n}}}-4}{\frac{4}{2n}}\right)$
So inside the exponential we have $\frac{0}{0}$ form and you can apply L'Hospital.
(just take $x=\frac{1}{n}$ and then evaluate at $x\to {0}^{+}$ for easier calculation.)
$\underset{x\to 0}{lim}\frac{3{e}^{\frac{-t\sqrt{x}}{4}}+{e}^{\frac{3t\sqrt{x}}{4}}-4}{2x}$
Using $x={h}^{2}$.(Continuous function) We have:-
$\underset{h\to 0}{lim}\frac{3{e}^{\frac{-th}{4}}+{e}^{\frac{3th}{4}}-4}{2{h}^{2}}$
Again this is $\frac{0}{0}$ form . So again L'Hospitaling it
$\underset{h\to 0}{lim}\frac{\frac{3{t}^{2}}{16}{e}^{\frac{-th}{4}}+\frac{9{t}^{2}}{16}{e}^{\frac{3th}{4}}}{4}=\frac{12{t}^{2}}{16\cdot 4}=\frac{3{t}^{2}}{16}$
So the original limit converges to
$\mathrm{exp}\left(\frac{3{t}^{2}}{16}\right)$
This just means that the MGF of the sequence of Rv's converges to the MGF of an $N\left(0,\frac{3}{8}\right)$
distribution.
Where $N\left(\mu ,{\sigma }^{2}\right)$ denotes the normal distribution with mean $\mu$ and standard deviation $\sigma$