How can I calculate this limit without L'Hospital rule and Taylor series?

$${\lim_{x \to 1} \big(4^x - 3^x\big)^{\frac{1}{x - 1}}}$$

  • $\begingroup$ Not sure if this is helpful, but the only thing I am thinking about is factoring out the $4$: $$ \begin{split} (4^x-3^x)^{1/(x-1)} &= 4 \left(4 - 4 \cdot \left(\frac{3}{4}\right)^x\right)^{1/(x-1)} \end{split} $$ $\endgroup$ – gt6989b Dec 29 '15 at 19:56
  • $\begingroup$ Just as a reference, Wolfram computes the limit as $\frac{256}{27}$. But what have you tried? $\endgroup$ – Moya Dec 29 '15 at 19:58
  • $\begingroup$ (This is for Clarinetist's deleted post, which was deleted while I was writing this comment. I think the deletion should be deleted.) @IgorRivin: I spent the last 20 minutes on this looking for another way (I also stayed away from any kind of derivative reformulation, thinking it might be solvable by precalculus methods), but I think Clarinetist is correct in thinking that using the definition of the derivative is probably fair game for this problem. $\endgroup$ – Dave L. Renfro Dec 29 '15 at 20:32
  • $\begingroup$ @DaveL.Renfro Deletion is deleted. $\endgroup$ – Clarinetist Dec 29 '15 at 21:17

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 $$\lim_{x \to 1}f(x) = \dfrac{256}{27}\text{.}$$

  • $\begingroup$ Looks an awful lot like l'hopital to me... $\endgroup$ – Igor Rivin Dec 29 '15 at 20:10
  • $\begingroup$ @IgorRivin Where did I use L'Hopital? $\endgroup$ – Clarinetist Dec 29 '15 at 20:11
  • $\begingroup$ You reduced the limit to a derivative computation, which is exactly how you would prove the general form of L'Hopital's rule... $\endgroup$ – Igor Rivin Dec 29 '15 at 20:12
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    $\begingroup$ @IgorRivin I think it's generally accepted that calculus students know the limit definition of the derivative and properties of limits before touching L-Hopital and Taylor series. For now, I will keep it up, but if someone posts a better answer, I will delete this. $\endgroup$ – Clarinetist Dec 29 '15 at 20:14
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    $\begingroup$ @IgorRivin Nowhere was a limit of a quotient of derivatives taken, hence L'Hopital was not used, and using the definition of the derivative alone is not how you prove L'Hopital. $\endgroup$ – zhw. Dec 29 '15 at 21:31

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

$$\lim_{x\to 1}\left(4^x-3^x\right)^{1/(x-1)}=\frac{256}{27}$$

  • $\begingroup$ I provided a way forward that is devoid of L'Hospital's Rule or other derivative-based approaches. Please let me know how I can improve my answer. I really want to give you the best answer I can. Happy Holidays. - Mark $\endgroup$ – Mark Viola Dec 30 '15 at 6:30
  • $\begingroup$ Nice...........+1 @Dr M.V. $\endgroup$ – Bhaskara-III Dec 30 '15 at 13:53
  • $\begingroup$ @bhaskara Thanks! Much appreciative. $\endgroup$ – Mark Viola Dec 30 '15 at 14:30

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