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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.

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  • $\begingroup$ Can you use the Taylor series? If yes the result is straightforward. $\endgroup$ – user63181 May 31 '14 at 12:38
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    $\begingroup$ @Tunk-Fey well, that does not sound very rigorous. $\endgroup$ – user66906 May 31 '14 at 12:40
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    $\begingroup$ No it's rigorous but it's simply the Taylor series. $\endgroup$ – user63181 May 31 '14 at 12:41
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    $\begingroup$ What Tunk-Fey wrote is strictly rigorous if we correct the typo (power $2/3$ should be $3/2$). $\endgroup$ – Claude Leibovici May 31 '14 at 12:46
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    $\begingroup$ How is it rigorous? I could say we can ignore the term $\dfrac a x$ since $\dfrac a x << 1$ as $x \to \infty$. Clearly this isn't the case though. $\endgroup$ – user85798 May 31 '14 at 13:14
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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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HINT:

$$\left(1+\frac ax+\frac b{x^{\frac32}}\right)^x=\left[\left(1+\frac{a\sqrt x+b}{x^{\frac32}}\right)^{\frac{x^{\frac32}}{a\sqrt x+b}}\right]^{\frac{a\sqrt x+b}{\sqrt x}}$$

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  • $\begingroup$ You write damn fast!! +1 $\endgroup$ – Paramanand Singh May 31 '14 at 12:44
  • $\begingroup$ @ParamanandSingh, It was really tough to write. But it was not right, at first :) $\endgroup$ – lab bhattacharjee May 31 '14 at 12:52
  • $\begingroup$ I noticed your mistake, but by the time i could tell about it, it was corrected...! $\endgroup$ – Paramanand Singh May 31 '14 at 12:58
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$$\lim_{x \rightarrow \infty} \left( 1+\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}} \right)^{\Large\frac{1}{\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}}}\cdot \left(\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}}\right)x}$$ 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 $$\lim_{x\to\infty}\left(\frac{a}{x}+\frac{b}{x^{\frac{3}{2}}}\right)x=a$$

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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 $$\left(1+\frac{1}{n/x_n}\right)^{n/x_n}\to e.$$

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