Limit problem with an exponential function without L'Hopital How can I calculate this limit without L'Hospital rule and Taylor series?
$${\lim_{x \to 1} \big(4^x - 3^x\big)^{\frac{1}{x - 1}}}$$
 A: Let $f$ be defined by $$f(x) = (4^x - 3^x)^{1/(x-1)}\text{.}$$
Define $g$ by $g = \ln f$. Then notice that
$$g(x) = \dfrac{\ln(4^x - 3^x)}{x-1}$$
Observe, furthermore, that if we set $h$ to be $$h(x) = \ln(4^x -3^x)$$
then 
$$h(1) = \ln(4-3) = \ln(1) = 0$$
so we have
$$\lim\limits_{x \to 1}g(x) = \lim\limits_{x \to 1}\dfrac{h(x)-h(1)}{x-1} = h^{\prime}(1)$$
(this is the limit definition of the derivative here!) and the derivative of $h$ is given by 
$$h^{\prime}(x) = \dfrac{1}{4^x - 3^x}[4^x \ln(4) - 3^x \ln(3)]$$
(recall that the derivative of $a^x$ is $a^x\ln(a)$) so 
$$h^{\prime}(1) = 4\ln(4)-3\ln(3)\text{.}$$
Thus we have shown
$$\lim_{x \to 1}g(x) = \lim_{x \to 1}\ln[f(x)] = 4\ln(4)-3\ln(3)\text{.}$$
For simplicity, rewrite
$$4\ln(4)-3\ln(3) = \ln(4^4)-\ln(3^3) = \ln(256) - \ln(27) = \ln\left(\dfrac{256}{27}\right)\text{.}$$
Hence, by continuity of $\ln$,
$$\lim_{x \to 1}g(x) = \ln\left(\dfrac{256}{27}\right) = \lim_{x \to 1}\ln[f(x)] = \ln\left[\lim_{x \to 1}f(x)\right]$$
and with
$$\ln\left[\lim_{x \to 1}f(x)\right] = \ln\left(\dfrac{256}{27}\right)$$
it follows that
$$\lim_{x \to 1}f(x) = \dfrac{256}{27}\text{.}$$
A: In THIS ANSWER, I showed using standard, non-calculus based analysis that 
$$\frac{x}{x+1}\le \log(1+x)\le x$$
for $x\ge -1$ and 
$$1+x\le e^x \le \frac{1}{1-x}$$
for $x<1$.
Now, using $(1)$ we can write
$$\frac{4^x-3^x-1}{(4^x-3^x)(x-1)}\le \frac{\log(4^x-3^x)}{x-1}\le \frac{4^x-3^x-1}{x-1}$$
Next, using $(2)$ we see that 
$$\begin{align}
\frac{4^x-3^x-1}{x-1}&=\frac{4(4^{x-1})-3(3^{x-1})-1}{x-1}\\\\
&=\frac{4e^{\log(4)(x-1)}-3e^{\log(3)(x-1)}-1}{x-1}\\\\
&\le \frac{\frac{4}{1-\log(4)(x-1)}-3(1+\log(3)(x-1))-1}{x-1}\\\\
&=\frac{4\log(4)-3\log(3)+3\log(3)\log(4)(x-1)}{1-\log(4)(x-1)}\\\\
&\to 4\log(4)-3\log(3)\,\,\text{as}\,\,x\to 1 \tag 3
\end{align}$$
We also see using $(2)$ that
$$\begin{align}
\frac{4^x-3^x-1}{(4^x-3^x)(x-1)}&=\frac{4(4^{x-1})-3(3^{x-1})-1}{(4^x-3^x)(x-1)} \\\\
&\ge \frac{4(1+\log(4)(x-1))-3\frac{1}{1-\log(3)(x-1)}-1}{(4^x-3^x)(x-1)}\\\\
&=\frac{4\log(4)-3\log(3)-4\log(3)\log(4)(x-1)}{(4^x-3^x)(1-\log(3)(x-1))}\\\\
&\to 4\log(4)-3\log(3)\,\,\text{as}\,\,x\to 1 \tag 4
\end{align}$$
We use $(3)$ and $(4)$ along with the Squeeze Theorem to reveal 
$$\lim_{x\to 1}\frac{\log(4^x-3^x)}{x-1}=\log\left(\frac{4^4}{3^3}\right)$$
Finally, we have
$$\lim_{x\to 1}\left(4^x-3^x\right)^{1/(x-1)}=\frac{256}{27}$$
