Limit of Moment Generating Function [closed]

Solution 1:

This has the $1^{\infty}$ form.

So you can use this $$\lim_{x\to a}f(x)^{g(x)} = \exp(\lim_{x\to a} (f(x)-1)g(x))$$ when $\lim_{x\to a}f(x)^{g(x)}$ is $1^{\infty}$ form.

Using the above we have :-

$$\exp(\lim_{n\to\infty}\frac{3e^{\frac{-t}{4\sqrt{n}}}+e^{\frac{3t}{4\sqrt{n}}}-4}{\frac{4}{2n}})$$

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.)

$$\lim_{x\to 0}\frac{3e^{\frac{-t\sqrt{x}}{4}}+e^{\frac{3t\sqrt{x}}{4}}-4}{2x}$$.

Using $x=h^{2}$.(Continuous function) We have:-

$$\lim_{h\to 0}\frac{3e^{\frac{-th}{4}}+e^{\frac{3th}{4}}-4}{2h^{2}}$$

Now using L'Hospital we have:-

$$\lim_{h\to 0}\frac{3\cdot\frac{-t}{4}e^{\frac{-th}{4}}+\cdot\frac{3t}{4}e^{\frac{3th}{4}}}{4h}$$.

Again this is $\frac{0}{0}$ form . So again L'Hospitaling it

$$\lim_{h\to 0}\frac{\frac{3t^{2}}{16}e^{\frac{-th}{4}}+\frac{9t^{2}}{16}e^{\frac{3th}{4}}}{4}=\frac{12t^{2}}{16\cdot 4} = \frac{3t^{2}}{16}$$.

So the original limit converges to $$\exp(\frac{3t^{2}}{16})$$.

This just means that the MGF of the sequence of Rv's converges to the MGF of an $N(0,\frac{3}{8})$ distribution.

Where $N(\mu,\sigma^{2})$ denotes the normal distribution with mean $\mu$ and standard deviation $\sigma$.