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I am attempting to show that

If $f(x) - g(x) \ll 1,\, x \to \infty$, then $e^{f(x)}\sim e^{g(x)}, \,x\to \infty$

From the first line, I am able to show that

$$ \lim_{x\to \infty} \frac{f(x) - g(x)}{1} = 0$$ from which it is clear that

$$ \lim_{x\to \infty}\left[f(x) - g(x)\right] = 0$$

However, I am a little stuck at this point. Where do exponentials come in?

Any help/hints much appreciated.

Simply Beautiful Art
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Victoria
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  • You can write $f(x)-g(x)\ll 1$; you don't have to write $f(x)-g(x)<<1$. Similarly, writing $$ e^{f(x)},\text{~},e^{g(x)}$$ is not proper MathJax usage; you should instead write $f(x)\sim g(x)$. $\qquad$ – Michael Hardy Mar 16 '16 at 05:28

1 Answers1

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As you've shown $$f(x)-g(x)\to 0,\quad\text{as}\, x\to\infty$$ So $$\frac{e^{f(x)}}{e^{g(x)}}=e^{f(x)-g(x)}\to 1$$ All you need is $$e^v\to 1,\quad\text{as}\,v\to 0$$

Vim
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