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$$\lim_{t\to 0}\ln\left(1+\frac{1}{t}\right)^t$$

Solved.

The existence and evaluation without LHospital rule.

user588102
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  • The limit exists, and it's equal to $0$. – Crostul May 23 '19 at 10:25
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    @Crostul But $\lim x\to 0^-1 \ne \lim x\to 0^+$ – user588102 May 23 '19 at 10:27
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    Your function is not defined in the region to the left however, so I would interpret $\lim_{x\to0}$ to mean $\lim_{x\to0^+}$. – John Doe May 23 '19 at 10:49
  • @JohnDoe But What if Lim LSH is undefined on real number set, can’t I say limit not exist? – user588102 May 23 '19 at 11:01
  • @user588102 I wouldn't say the limit does not exist since the function itself isn't defined on the left. It doesn't make sense to see what happens to the function as you approach $0$ from the left, since what would the value of the function be at, say $-0.1$? What about as you get closer to $0$? This issue is more with the definition of the function than anything to do with limits. – John Doe May 23 '19 at 11:05
  • @JohnDoe I think I now got your point. So it is possible to find the limit without L’hospital rule? – user588102 May 23 '19 at 11:11

2 Answers2

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Your function is not defined in the region to the left however, so I would interpret $\lim_{x\to0}$ to mean $\lim_{x\to0^+}$.

If you want to define it in this region, you can choose the branch $\ln\left(-x\right)=\ln(x)+i\pi$. ($x>0$). Then

$$\lim_{x\to0^-}\left\{x\ln\left(1+\frac1x\right)\right\}=\lim_{x\to0^-}\left\{-|x|\ln\left(\frac1{|x|}-1\right)-i\pi|x|\right\}=\cdots=0$$

So the limit can also be shown to be $0$ from the left after defining the function in that region.

John Doe
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  • From my knowledge, i only know the definition of log on real numbers, so that mean it is not define on real number on the LHS of 0. And the definition of limit of a function is LHS limit=RHS limit, so if LHS is undefined, can I say limit doesn’t exist? – user588102 May 23 '19 at 11:05
  • LHS limit is not undefined. It's simply not a thing you can work out. When you work out a limit, you are supposed to approach the limiting point ($x_0$)but choosing closer and closer points to $x_0$ in the domain of the function. You cannot get arbitrarily close to $x_0=0$ in this example, since the function isn't defined in the range $x\in[-1,0]$, so these points aren't in the domain of the function. – John Doe May 23 '19 at 11:14
  • Thank you very much, I think I have a better understand to limit now and I understand what you mean. My second questions is whether it could be found without Lhospital rule for finding $\lim 0\to 0$? – user588102 May 23 '19 at 11:18
  • @user588102 that, I am not sure about. You could write it as $$-x\ln x +x\ln(1+x)$$ which eliminates the second term. But I don't know if it is possible to evaluate $\lim_{x\to 0} x\ln x$ without L'Hopital's rule. Its a limit that I'd always prefer to use L'Hopital's for, since it simplifies so nicely... – John Doe May 23 '19 at 11:26
  • @user588102 see this answer. It evaluates $x^x$ as $x\to0$ without L'Hopital, and the limit you need to work out is $x\ln x=\ln(x^x)$. Since the limit is shown to be $1$, this means your limit is $\ln(1)=0$. – John Doe May 23 '19 at 11:33
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For sanity, replace $\frac{1}{x} = t$ and note that as $x \to 0, t \to \infty$. So our limit is

$$\lim_{t\to \infty} \frac{ \log(1+t)}{t}$$

Now make the substitution,

$$\log(1+t) = y$$

$$\implies t = e^y -1$$

Also note that as $t\to \infty$, $y \to \infty$. Therefore our limit is:

$$\lim_{y \to \infty} \frac{y}{e^y -1}$$

Expanding $e^y$,

$$e^y = 1+y + \frac{y^2}{2!} + ... $$

We get,

$$\lim_{y \to \infty} \frac{y}{y+\frac{y^2}{2!} + ... }$$

$$= \lim_{y\to\infty} \frac{1}{1+\frac{y}{2!} + ... }$$

$$=0$$

Edit: We are only concerned with $x \to 0^+$ in the original limit as the function is not defined at $x \to 0^-$.

Vizag
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