calculus - 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.

Answer

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 $L = e^{a}$.