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 ?

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    $\begingroup$ 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 $c = 0$. $\endgroup$ – Paramanand Singh Apr 5 '14 at 4:48

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 $$\lim_{x\rightarrow x_0}{f(x)^{g(x)}}=\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}$$

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  • $\begingroup$ The limit might now exist at the left of $x=x_0$.Then how do I assume that the log goes inside? $\endgroup$ – user586228 Oct 14 at 5:47


  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

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

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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 $f(x_n,y_n)\to f(x,y)$.

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