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Arthur Mattuck in his Introduction to Analysis book, pg. 220 says, in order to prove L'Hospital's Rule for $\infty/\infty$ case,

Let $L=\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ and choose $a$ so that

$\frac{f'(x)}{g'(x)} \stackrel{\approx}{\epsilon} L$ for $x>a$.

And prove two approximations below (valid for $x\gg 1$)

$$ \frac{f(x)}{g(x)} \stackrel{\approx}{\epsilon} \frac{f(x)-f(a)}{g(x)-g(a)} \stackrel{\approx}{\epsilon} L $$

His hint is, for the first approximation, that we write

$$ f(x) - f(a) = f(x) [ 1 - f(a)/f(x)] $$

and use limit theorem, for the second, use the Cauchy Mean-value Theorem.

Anyone know how to pursue this proof? Thanks,

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Could you please explain the notation $\stackrel{\approx}{\epsilon}$? – Brad Dec 23 '11 at 17:27
My guess is that it means $|f'(x)/g(x) - L| < \epsilon$. I agree that it isn't very standard and should be explained. – Dylan Moreland Dec 23 '11 at 17:29
To the OP: Do you mean to write $f'(x)/g'(x)$? – Dylan Moreland Dec 23 '11 at 17:29
Brad, if $a$ and $b$ are within $\epsilon$ of eachother, then Mattuck uses $a \stackrel{\approx}{\epsilon} b$. – BB_ML Dec 23 '11 at 18:08
Dylan, correction, yes the first ratio is between derivatives. – BB_ML Dec 23 '11 at 18:19
up vote 4 down vote accepted

The first approximation should be clear since we can, for any $\epsilon$, choose $x_0$ sufficiently large such that $$1 - \epsilon < \frac{1 - f(a)/f(x)}{1-g(a)/g(x)} < 1 + \epsilon$$ whenever $x > x_0$.

For the second approximation: by the Cauchy MVT, there exists $c, a < c < x$ such that $(f(x) - f(a))g'(c) = (g(x)-g(a))f'(c)$, that is: $$\frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c)}{g'(c)}$$ which is $\epsilon$-close to $L$.

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