# Does this limit hold?

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

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

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

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

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.

• The condition $l_2\neq 0$ suffices to assure that eventually $g(x)\neq 0$.
– user
Sep 30, 2020 at 7:40

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 $$g$$.