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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.

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    $\begingroup$ The limit to $0$ does exist, because limits are computed on the domain only (and the limit to $0^-$ is irrelevant). $\endgroup$
    – user65203
    Sep 4, 2017 at 17:14
  • $\begingroup$ 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 $x<0$. $\endgroup$
    – user296113
    Sep 4, 2017 at 17:17
  • $\begingroup$ @YvesDaoust Can you explain more about that property I didn't know? (Or link me to something) $\endgroup$
    – Winter
    Sep 4, 2017 at 17:19
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    $\begingroup$ Review your definition of a limit. $\endgroup$
    – user65203
    Sep 4, 2017 at 17:20

2 Answers 2

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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 $x<x_0$ 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.

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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 $$\frac{\ln (\tan 2x)}{\ln (\tan 3x)}\sim_0\frac{\ln 2x}{\ln 3x}=\frac{\ln 2+\ln x}{\ln 3+\ln x}\xrightarrow[x\to0^+]{}1.$$

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

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