How do I show that $\lim_{x \rightarrow \infty}(1+\frac{a}{x}+\frac{b}{x^{3/2}})^x =e^{a}$? How do I show that $\lim_{x \rightarrow \infty}(1+\frac{a}{x}+\frac{b}{x^{3/2}})^x  = e^a$? Actually, I had to deal with something similar yesterday and after thinking about it for quite a while I did it with L'Hospital's rule, but this was very unsatisfactory for me. I am rather interested in a more algebraic proof that this last term does not contribute to the limit, but I found it quite hard to do something more elementary.
 A: I am expanding the hint in comments. In dealing with limits of expression of type $\{f(x)\}^{g(x)}$ it is much better to take logs rather than write complicated exponents. Let the limit be $L$. Then we have $$\begin{aligned}\log L &= \log\left\{\lim_{x \to \infty}\left(1 + \frac{a}{x} + \frac{b}{x^{3/2}}\right)^{x}\right\}\\
&=\lim_{x \to \infty}\log\left(1 + \frac{a}{x} + \frac{b}{x^{3/2}}\right)^{x}\text{ because log is continuous}\\
&= \lim_{x \to \infty}x\log\left(1 + \frac{a}{x} + \frac{b}{x^{3/2}}\right)\\
&= \lim_{x \to \infty}x\cdot\left(\dfrac{a}{x} + \dfrac{b}{x^{3/2}}\right)\dfrac{\log\left(1 + \dfrac{a}{x} + \dfrac{b}{x^{3/2}}\right)}{\dfrac{a}{x} + \dfrac{b}{x^{3/2}}}\\
&= \lim_{x \to \infty}\left(a + \frac{b}{\sqrt{x}}\right)\cdot \lim_{y \to 0}\frac{\log(1 + y)}{y}\text{ where }y = \frac{a}{x} + \frac{b}{x^{3/2}}\\
&= a\cdot 1 = 1\end{aligned}$$ Hence $L = e^{a}$.
A: HINT:
$$\left(1+\frac ax+\frac b{x^{\frac32}}\right)^x=\left[\left(1+\frac{a\sqrt x+b}{x^{\frac32}}\right)^{\frac{x^{\frac32}}{a\sqrt x+b}}\right]^{\frac{a\sqrt x+b}{\sqrt x}}$$
A: $$\lim_{x \rightarrow \infty} \left( 1+\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}} \right)^{\Large\frac{1}{\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}}}\cdot \left(\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}}\right)x}$$
The term
$$\lim_{x \rightarrow \infty} \left( 1+\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}} \right)^{\Large\frac{1}{\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}}}} = e$$
and 
$$\lim_{x\to\infty}\left(\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}}\right)x=a$$
A: More generally, one can show that if
$x_n \to x$
then 
$$e^x=\lim\limits_{n \to\infty}  \left(1+\frac{x_n}{n}\right)^n.$$
by the same ideas as the other answers, as follows
$$\left(1+\frac{x_n}{n}\right)^n= \left[ 
\left(1+\frac{1}{n/x_n}\right)^{n/x_n}\right]^{x_n}$$
And $$\left(1+\frac{1}{n/x_n}\right)^{n/x_n}\to e.$$
