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 $L = e^{a}$.
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