Solving $\lim_{x\to1}(4^x-3^x)^{\frac{1}{1-x}}$ and others like it without L'hospital Ok so I have issues with this specific 'type' of limits. : $\\$ 
$$\lim_{x\to1}(4^x-3^x)^{\frac{1}{1-x}}$$
and $\\$
$$\lim_{x\to-1}\biggr(-4\cdot\arctan(x)\cdot{\frac{1}{\pi}}\biggr)^{\frac{1}{x+1}}$$
$\\$
It seems like they are quite similar, but I'm not sure what to do. I've tried taking $\ln$ of the limit to simplify it but have reached nothing. I would appreciate any hint whatsoever.
However, I am not allowed to use L'hospital rule or integrals to solve these.
Thanks in advance!
 A: Change the variable to $x=y+1$ and you have now $y\to 0$.
Next take the log so you have 
$$-\frac{\ln(4^{y+1}-3^{y+1})}{y}$$
Now use that $$\frac{\ln(4^{y+1}-3^{y+1})}{4^{y+1}-3^{y+1}-1}\to 1$$ as a special instance of $\lim\limits_{t\to 0}\frac{\ln 1+t}{t}=1$.
So the problem reduces to 
$$-\frac{4^{y+1}-3^{y+1}-1}{y}$$
which we write as 
$$\frac{3^{y+1}-3}{y}-\frac{4^{y+1}-4}{y}=3\frac{3^{y}-1}{y}-4\frac{4^{y}-1}{y}$$
Finally we use the fact that $$\frac{a^t-1}{t}\to \ln a$$ to see that your limit is 
$$3\ln 3-4\ln 4$$
But dont forget to undo the log so the real limit is $$\frac{3^3}{4^4}$$
A: I thought it might be instructive to present an approach that relies only on a set of elementary inequalities.  To that end, we begin with a primer.


PRIMER:
In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that the exponential function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$
for $x<1$, and the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag2$$
for $x>0$


We note that without loss of generality, we assume that the limit is taken from the left-hand side (i.e., $x<1$).  Proceeding, let $f(x)=(4^x-3^x)^{1/(1-x)}$.  Then, using $t=e^{\log(t)}$ we can write $f(x)$ as
$$\begin{align}
f(x)=e^{\frac{\log(4^x-3^x)}{1-x}} \tag 3
\end{align}$$
Letting $g(x)=\frac{\log(4^x-3^x)}{1-x}$ and exploiting $(1)$ and $(2)$, the upper bound for $g(x)$ is 
$$\begin{align}
g(x)&\le \frac{4^x-3^x-1}{1-x}\\\\
&=\frac{4e^{(x-1)\log(4)}-3e^{(x-1)\log(3)}-1}{1-x}\\\\
&\le \frac{\frac{4}{1-(x-1)\log(4)}-3(1+(x-1)\log(3))-1}{1-x}\\\\
&=\frac{3\log(3)-4\log(4)-3\log(3)\log(4)(x-1)}{(1-(x-1)\log(4))}\\\\
&\to \log\left(\frac{3^3}{4^4}\right)\,\,\,\text{as}\,\,x\to 1 \tag 4
\end{align}$$
and the lower bound for $g(x)$ is 
$$\begin{align}
g(x)&\ge \frac{4^x-3^x-1}{(1-x)(4^x-3^x)}\\\\
&=\frac{4e^{(x-1)\log(4)}-3e^{(x-1)\log(3)}-1}{(1-x)(4^x-3^x)}\\\\
&\ge \frac{4(1+(x-1)\log(4))-\frac{3}{1-(x-1)\log(3)}-1}{(1-x)(4^x-3^x)}\\\\
&=\frac{3\log(3)-4\log(4)+4\log(3)\log(4)(x-1)}{4^x-3^x}\\\\
&\to \log\left(\frac{3^3}{4^4}\right)\,\,\,\text{as}\,\,x\to 1 \tag 5
\end{align}$$
Applying the squeeze theorem to $(4)$ and $(5)$ shows that $\lim_{x\to 1}g(x)=\log\left(\frac{3^3}{4^4}\right)$ whereupon using $(3)$ and exploiting the continuity of the exponential function yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1}(4^x-3^x)^{1/(1-x)}}=\frac{3^3}{4^4}$$

Note that the only tools used were the inequalities in $(1)$ and $(2)$ along with the continuity of the exponential function and the squeeze theorem!
