If $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}g(x)=0$, and $\lim_{x \to \infty}\frac{f'(x)}{g'(x)}=l$, then $\lim_{x \to \infty}\frac{f(x)}{g(x)}=l$

Solution 1:

There are three lemmas that we will need:

If $f$ is continuous on an interval and for any $a,b$ in the interval $f(a) \neq f(b)$, then $f$ is strictly increasing or strictly decreasing on that interval $\quad (\dagger_1)$

If $\displaystyle \lim_{x \to \infty}g(x)=0$ and there is an $N$ such that on the interval $(N, \infty)$ $g$ is strictly increasing or strictly decreasing, then $g(x) \neq 0$ for any $x \in (N, \infty). \quad (\dagger_2)$

If $\displaystyle \lim_{x \to \infty}f(x)=0$, $\displaystyle \lim_{x \to \infty}g(x)=0$, and there is an $N$ such that for any $x_1, x_2 \in (N, \infty)$, $g(x_1) \neq 0$ and $g(x_1)\neq g(x_2)$, then $\displaystyle \lim_{x \to \infty}\frac{f(c)-f(x)}{g(c)-g(x)}=\frac{f(c)}{g(c)}$ for any $c \in (N, \infty)$ $\quad (\dagger_3)$

Because $\displaystyle \lim_{x \to \infty}\frac{f'(x)}{g'(x)}=l$, we know that there is an $N_1$ such that:

$\forall x \left[x \gt N_1 \rightarrow \left|\frac{f'(x)}{g'(x)}-l \right|\lt \frac{\varepsilon}{2} \right]$

Clearly, then, for any $x \in (N_1, \infty)$, $\frac{f'(x)}{g'(x)}$ is defined. This means that for any $x \in (N_1, \infty)$, $g'(x) \neq 0$.

Additionally, because $g'(x)$ is necessarily defined on this interval (and is therefore continuous, as well, on this interval), we can apply Rolle's Theorem on any arbitrary interval $[x_1,x_2]$ where $x_1, x_2 \in (N_1, \infty)$ to assert that $g(x_1) \neq g(x_2)$. Otherwise, there would be an $x' \in (N,\infty)$ such that $g'(x')=0$, a contradiction. Thus, we can invoke $(\dagger_1)$ to say that on the interval $(N_1,\infty)$, $g$ is either strictly increasing or strictly decreasing.

Now, for any $x_1,x_2 \in (N_1,\infty)$, we can apply the Cauchy Mean Value Theorem on the interval $[x_1,x_2]$ to show that there is an $x^* \in (x_1,x_2)$ such that:

$$\left[f(x_2)-f(x_1) \right]g'(x^*) = \left[g(x_2)-g(x_1)\right]f'(x^*)$$

Because $x_1,x^*,x_2 \in (N_1,\infty)$, we can rearrange this to:

$$\frac{f(x_2)-f(x_1)}{g(x_2)-g(x_1)}=\frac{f'(x^*)}{g'(x^*)}$$

We can combine this with our statement involving $\frac{\varepsilon}{2}$ to assert that, for any $x_1, x_2 \in (N_1, \infty)$:

$$\left|\frac{f(x_2)-f(x_1)}{g(x_2)-g(x_1)}-l \right| \lt \frac{\varepsilon}{2} \quad (\star_1)$$

By assumption, $\displaystyle \lim_{x \to \infty}g(x)=0$. Further, we previously showed that $g$ is strictly increasing or strictly decreasing on $(N_1,\infty)$. Therefore, we can invoke $(\dagger_2)$, which says that for any $x \in (N_1,\infty)$, $g(x)\neq 0$.

We are in position to invoke $(\dagger_3)$. In particular, choose an instance of $\frac{\varepsilon}{2}$ to produce the statement:

for any $c \gt N_2$: $\forall x \left[x \gt N_2 \rightarrow \left|\frac{f(c)-f(x)}{g(c)-g(x)} - \frac{f(c)}{g(c)} \right| \lt \frac{\varepsilon}{2}\right] \quad (\star_2)$

Finally, consider the $N=\max(N_1,N_2)$. Let $c$ and $x$ be any arbitrary element in $(N,\infty)$.

Using $(\star_1)$ and $(\star_2)$, we can say:

$\left|\frac{f(c)}{g(c)} - \frac{f(c)-f(x)}{g(c)-g(x)} \right| \lt \frac{\varepsilon}{2}$

and

$\left |\frac{f(c)-f(x)}{g(c)-g(x)}-l \right| \lt \frac{\varepsilon}{2}$

Recalling the triangle equality rule applied in the following manner:

$|(a-b)-(b-c)| \leq |a-b| + |b-c|$

we can combine the above two statements to produce:

$$\left|\frac{f(c)}{g(c)}-l \right| \lt \varepsilon$$

However, note that $c$ was any arbitrary element in $(N,\infty)$. Therefore we can generalize, which means we have shown that for any $\varepsilon$, we can construct an $N$ such that for any $x \gt N$: $\left|\frac{f(x)}{g(x)}-l \right| \lt \varepsilon \iff \displaystyle \lim_{x \to \infty}\frac{f(x)}{g(x)}=l$