Solving $\lim_{x\to1}(4^x-3^x)^{\frac{1}{1-x}}$ and others like it without L'hospital

Question

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!

Answer 1

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 $$\frac{3^3}{4^4}$$

Answer 2

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.

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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 $x>0$

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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!