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I want to know the following is true : If $$ c,\ d\in {\bf R},\ \lim_{x\rightarrow 0} f(x)=c>0,\ \lim_{x\rightarrow 0} g(x) =d>0$$ then $$ \lim_{x\rightarrow 0} f(x)^{g(x)} = c^d$$

In calculus book such formula cannot be found.

Consider the problem : $$\lim_{x\rightarrow 0} (1+\sin\ 4x)^{\cot\ x} $$

To find a limit, we must use ${\rm log}$ and L'Hospital. But some student suggests that $$ \lim_{x\rightarrow 0} (1+\sin\ 4x)^{\frac{1}{\sin\ 4x} \frac{\sin\ 4x}{\sin\ x}\cos\ x}=e^4$$

This argument is clear. But I know that it is informal. Is there a minus point in such way ?

HK Lee
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    The result you have mentioned is true (as shown in the answer) but it is much better to take logs and then evaluate the logarithm. The formula is valid for all $c > 0$ and any finite value of $d$. But the technique of taking logs works in other cases too one of $c, d$ is infinite or if $c = 0$. – Paramanand Singh Apr 05 '14 at 04:48
  • @ParamanandSingh Here how can we apply $\ln$ without knowing inner limit exist? – Meet Patel May 11 '23 at 03:18
  • @MeetPatel: this works because $\ln$ is a continuous bijection. The interplay of $\ln$ and its inverse $\exp $ is needed here. See more details at https://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-3.html?m=0 – Paramanand Singh May 11 '23 at 04:22

3 Answers3

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Assuming $f(x)>0$ for some open set surrounding $x_0$, we know that because the natural logarithm is continuous on $(0,\infty)$ that \begin{align} \ln \left( \lim_{x\rightarrow x_0}{f(x)^{g(x)}}\right) & =\lim_{x\rightarrow x_0}{\ln \left(f(x)^{g(x)}\right)} \\ & =\lim_{x\rightarrow x_0}{g(x)\ln f(x)} \\ & =\left(\lim_{x\rightarrow x_0}{g(x)} \right)\left(\lim_{x\rightarrow x_0}{\ln f(x)}\right) \\ & =\left(\lim_{x\rightarrow x_0}{g(x)}\right)\ln \left( \lim_{x\rightarrow x_0}{f(x)}\right) \\ & =\ln \left[\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}\right] \end{align}

Then, because the natural logarithm is injective, we find that $$\lim_{x\rightarrow x_0}{f(x)^{g(x)}}=\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}$$

Hayden
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Yes. Note that the function $f(x,y)=x^y=e^{y\ln(x)}$ is continuous for $x,y>0$, so if $(x_n,y_n)\to (x,y)$ with $x,y>0$ then $f(x_n,y_n)\to f(x,y)$.

Alex Becker
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Because

  1. $y(x)=ln(x)$ is continuous at $x = c > 0$

  2. $\lim\limits_{x\to 0}f(x) = c$

according to the composition law, we have

$$\lim\limits_{x \to 0}lnf(x) = ln\lim\limits_{x \to 0}f(x) = lnc$$

Because $\lim\limits_{x \to 0}g(x) = d$, we have

$$\lim\limits_{x\to 0}g(x)lnf(x) = \lim\limits_{x\to 0}g(x)\cdot\lim\limits_{x \to 0}lnf(x) = dlnc$$

Apply composition law again, we get

$$\lim\limits_{x\to 0}f(x)^{g(x)} = \lim\limits_{x\to 0}e^{g(x)lnf(x)} = e^{\lim\limits_{x\to 0}g(x)lnf(x)} = e^{dlnc} = c^d$$

Searene
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