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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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    Dylan, correction, yes the first ratio is between derivatives.2011-12-23

1 Answers 1

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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 $L$.