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Let us suppose that there is a function $f(x)$ of the form $g(x)/h(x)$. When limit as $x \to \infty$ cannot be determined outright, one can use l'hopital's rule. Suppose that by performing L'hopital's rule, limit can be determined. But before limit is determined, we obtain $g'(x)/h'(x)$. Suppose that $f(x)$ has another form $i(x)/j(x)$. Again, limit cannot be determined outright, so we perform L'hopital's rule. Then we will get $i'(x)/j'(x)$. Again, let's say this derivative form allows us to calculate limit as $x \to \infty$. Then,

Would $g'(x)/h'(x) = i'(x)/j'(x)$?

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l'Hospital's rule is applieable only if you have an indeterminate form, $\;\frac00\;,\;\frac\infty\infty\;$ –  DonAntonio Oct 6 '13 at 19:28
    
Yeah, that's why I said limit was not able to be determined outright. –  Calc 101 Oct 6 '13 at 19:29
    
It's not the same, not even close: the limit must be an indeterminate of the mentioned forms or else you can't even try to apply l'H. –  DonAntonio Oct 6 '13 at 19:30

1 Answer 1

If $$f(x) = \frac{g(x)}{h(x)} = \frac{i(x)}{j(x)}$$ for all sufficiently large $x$ and $$\lim_{x\to\infty} g(x) = \lim_{x\to\infty} h(x) \in \{0,\pm \infty\}$$ and $$\lim_{x\to\infty} i(x) = \lim_{x\to\infty} j (x) \in \{0,\pm \infty\}$$ and $$\lim_{x\to\infty} \frac{g'(x)}{h'(x)},\qquad \lim_{x\to\infty} \frac{i'(x)}{j'(x)}$$ both exist, then $$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{g'(x)}{h'(x)} = \lim_{x\to\infty} \frac{i'(x)}{j'(x)}.$$

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