# Can we still use L'Hospital's rule on limit that does not exists?

I have the following limit and I must find the value of it:

$\lim\limits_{x\to0}{\frac{\ln(\tan 2x)}{\ln(\tan 3x)}}$

The corrected version of this exercise provided by my teacher says it equals 1 and that's the value I got using L'Hospital's rule.

However, from what I know of limits, the limit does not seem to exists since $\lim\limits_{x\to{0^+}{}}{{\ln(x)}}$ exists but not $\lim\limits_{x\to{0^-}{}}{{\ln(x)}}$. Thus $\lim\limits_{x\to{0}{}}{{\ln(x)}}$ does not exists either.

Am I wrong or is the corrected version wrong? Also, I tried to use online websites to find the limit's value and they also find 1 as well for the first limit as for the one I used as an example.

• The limit to $0$ does exist, because limits are computed on the domain only (and the limit to $0^-$ is irrelevant).
– user65203
Sep 4, 2017 at 17:14
• Even it isn't explicitly mentioned, the given limit is understood to be $\lim_{x\to 0^+}$ because the given function isn't defined for $x<0$. Sep 4, 2017 at 17:17
• @YvesDaoust Can you explain more about that property I didn't know? (Or link me to something) Sep 4, 2017 at 17:19
• Review your definition of a limit.
– user65203
Sep 4, 2017 at 17:20

## 2 Answers

If a function $$f$$ is a real-valued function whose domain is some subset $$A$$ of the real line for which $$x_0$$ is a limit point of both $$(-\infty,x_0)\cap A$$ and $$(x_0,\infty)\cap A,$$ then we can conclude that $$\lim_{x\to x_0} f(x)$$ exists if and only if $$\lim_{x\to x_0^-}f(x)$$ and $$\lim_{x\to x_0^+}f(x)$$ exist and are equal to each other.

In general, if we define limits as follows:

Suppose $$f$$ is a real-valued function whose domain is some subset $$A$$ of the real line, and suppose $$x_0\in\Bbb R.$$ We say the limit of $$f(x)$$ as $$x$$ approaches $$x_0$$ exists if (and only if) the following hold:

• $$x_0$$ is a limit point of $$A,$$ and

• there is some $$L\in\Bbb R$$ such that, for any $$\epsilon>0$$ (no matter how small), we can find some $$\delta>0$$ such that whenever $$x\neq x_0$$ is within $$\delta$$ of $$x_0,$$ we have $$f(x)$$ within $$\epsilon$$ of $$L.$$

We denote this $$L$$ (if it exists) by $$\lim_{x\to x_0}f(x).$$

As for one-sided limits, we define them as follows:

Suppose $$f$$ is a real-valued function whose domain is some subset $$A$$ of the real line, and suppose $$x_0\in\Bbb R.$$ We say the limit of $$f(x)$$ as $$x$$ approaches $$x_0$$ from the left exists if (and only if) the following hold:

• $$x_0$$ is a limit point of $$A\cap(-\infty,x_0),$$ and

• there is some $$L\in\Bbb R$$ such that, for any $$\epsilon>0$$ (no matter how small), we can find some $$\delta>0$$ such that whenever $$x and $$x$$ is within $$\delta$$ of $$x_0,$$ we have $$f(x)$$ within $$\epsilon$$ of $$L.$$

We denote this $$L$$ (if it exists) by $$\lim_{x\to x_0^-}f(x).$$ We similarly define the existence of the limit of $$f(x)$$ as $$x$$ approaches $$x_0$$ from the right, and denote it (if it exists) by $$\lim_{x\to x_0^+}f(x).$$

We can abuse the notations above to indicate increase/decrease without bound as $$x$$ approaches $$x_0$$ from two sides or just one. If $$x_0$$ is a limit point of both $$A\cap(-\infty,x_0)$$ and $$A\cap(x_0,\infty),$$ then $$f(x)$$ increases (decreases) without bound as $$x$$ approaches $$x_0$$ if and only if it does so as $$x$$ approaches $$x_0$$ from both the left and the right. However, if $$x$$ is only a limit point of either $$A\cap (-\infty,x_0)$$ or $$A\cap(x_0,\infty),$$ we can still talk about $$\lim_{x\to x_0}f(x),$$ as long as we can talk about the one-sided limit that actually makes sense.

In your particular cases, $$\lim_{x\to x_0^+}f(x)=\lim_{x\to x_0}f(x),$$ since neither numerator nor denominator is defined for $$x\le0.$$ So, you can still apply the rule.

The limit exists because, as pointed by Yves Daoust, $x$ belongs to the domain of the function, and in a neighbourhood of $0$, this implies $x>0$.

Anyway it's so much simpler to use equivalents! We know $\tan u\sim_0 u$, so $$\frac{\ln (\tan 2x)}{\ln (\tan 3x)}\sim_0\frac{\ln 2x}{\ln 3x}=\frac{\ln 2+\ln x}{\ln 3+\ln x}\xrightarrow[x\to0^+]{}1.$$

• You can't apply the equivalence "inside of" a function. Dec 20, 2023 at 12:14