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Suppose that $f$ and $g$ are continuously differentiable that converge to $l_1$ and $l_2$, when $x\to\infty$. Does it hold that

$$\lim_{x\to\infty}\left(\frac{f(x)}{g(x)}\right)^2=\left(\lim_{x\to\infty}\frac{f(x)}{g(x)}\right)^2=\left(\frac{l_1}{l_2}\right)^2$$

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  • $\begingroup$ Yes, if $l_2 \neq 0$. $\endgroup$ Sep 30, 2020 at 7:32

3 Answers 3

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$$\lim_{x\to\infty}\left(\frac{f(x)}{g(x)}\right)^2=\left(\lim_{x\to\infty}\frac{f(x)}{g(x)}\right)^2$$ holds by continuity of the square function.

Then

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\frac{\lim_{x\to\infty}f(x)}{\lim_{x\to\infty}g(x)}=\frac{l_1}{l_2}$$ holds by continuity of the division (if you prefer, by the division rule for limits), provided $l_2\ne0$.

Continuity and differentiability of $f,g$ play no role here.

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Provided that $l_2\neq 0$, then yes. In fact, if $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}g(x)$ both exist and satisfy the requirements above, this is true even if $f$ and $g$ are discontinuous.

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    $\begingroup$ The condition $l_2\neq 0$ suffices to assure that eventually $g(x)\neq 0$. $\endgroup$
    – user
    Sep 30, 2020 at 7:40
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Yes it holds under the condition that $l_1,l_2\in \mathbb R$ with $l_2\neq 0$.

Note also that continuity is not a necessary condition for $f$ and $g$.

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