Explanation for limits equality.

Question

$$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} \right)^{\frac{1}{x}}} = \exp \left( {\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{a^x} + {b^x}}}{2} - 1}}{x}} \right)$$

I am familiar with the "taking the exponent trick" which bring us to:

$$\\exp \left( {\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} + {b^x}}}{2}} \right) \cdot \frac{1}{x}} \right) $$

I truly don't get it. I'll be glad for an explanation.
Cheers!

Clarification: $exp(A) = e^A$.

Answer 1

\begin{align*} \left(\frac{a^x+b^x}{2}\right)^{\frac 1 x}=\exp\left(\frac 1 x{\color{\red}\ln}\left(\frac{a^x+b^x}{2}\right)\right). \end{align*}

Now, as $x\to0$, $(a^x+b^x)/2\to (a^0+b^0)/2=(1+1)/2=1$ (assuming that $a,b\neq0$). Also, $\ln(1)=0=1-1$, so that $\lim_{y\to1}\ln y=\lim_{y\to1}(y-1)$, since both functions are continuous at 1. Let $y\equiv(a^x+b^x)/2$. As have seen, $y\to1$ as $x\to0$, so that we have $$\lim_{x\to0}\ln\left(\frac{a^x+b^x}{2}\right)=\lim_{y\to1}\ln y=0=\lim_{y\to1}(y-1)=\lim_{x\to0}\left(\frac{a^x+b^x}{2}-1\right).$$

Edit: My previous solution contained an error, as pointed out by @AndréNicolas. I apologize. This is a revised version.

From here, we have that (pulling the limit inside is possible, since $x\mapsto\exp(x)$ is continuous) \begin{align*} =&\,\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{\frac 1 x}=\lim_{x\to0}\exp\left(\frac 1 x{\color{\red}\ln}\left(\frac{a^x+b^x}{2}\right)\right)=\exp\left(\lim_{x\to0}\frac1x\ln\left(\frac{a^x+b^x}{2}\right)\right)\\ =&\,\exp\left(\lim_{x\to0}\frac1 x\left(\frac{a^x+b^x}{2}-1\right)\right), \end{align*} provided that the two limits in the parentheses in the last two expressions are equal. But this is not too difficult to show. We can use L'Hôpital's rule, since both $\ln((a^x+b^x)/2)$ and $(a^x+b^2)/2-1$ converge to $0$ as $x\to 0$, as shown above, and the denominator $x$ naturally converges to zero in a sufficiently smooth way in each case, so that L'Hôpital's rule is applicable. I leave it as an exercise to show that the limit is $(\ln a+\ln b)/2$ in both cases, provided that $a,b>0$. (Hint: Use the fact that $(a^x)'=a^x\cdot\ln a$ for any $a>0$.)

Answer 2

Going to $$\lim_{x\to 0} \frac{\ln\left(\frac{a^x+b^x}{2}\right)}{x}\tag{1}$$ then presents no problem for you.

For brevity, let $t=f(x)=\frac{a^x+b^x}{2}-1$. Note that when $x$ is near $0$, then $f(x)$ is near $0$. Rewrite $\ln(1+f(x))$ (when $f(x)\ne 0$) as $$f(x) \frac{\ln(1+f(x))}{f(x)}.$$ So to find (1) we need to find $$\lim_{x\to 0}\frac{f(x)}{x}\cdot \frac{\ln(1+f(x))}{f(x)}.$$ The limit as $t\to 0$ of $\frac{\ln(1+t)}{t}$ is $1$, which gives the desired result.

Remark: We are basically expanding $\ln(1+f(x))$ in a power series (first two terms only).

Answer 3

Hint : In limits of the form $$\lim_{x\to a} {(f(x))^{g(x)}}$$ if it turns out that $\lim_{x\to a} {f(x)}=1$ and $\lim_{x\to a} {g(x)}=\infty$, then the following transformation may works:

$$\begin{align*} \lim_{x\to a} {(f(x))^{g(x)}} &=\lim_{x\to a} {(1+(f(x)-1))^{g(x)}}\\ &=\lim_{x\to a} {\{(1+(f(x)-1))^{1/(f(x)-1)}\}^{g(x)(f(x)-1)}}\\ &= e^{\lim_{x\to a}{g(x)(f(x)-1)}}\end{align*}$$

Now in your example $f(x)=\frac{{{a^x} + {b^x}}}{2}$ and $g(x)=\frac{1}{x}$, so...

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I hope it helps.