Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$

Question

Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$

Here is what I have done so far: \begin{align} \lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2} &= \lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\ &=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)\\ &=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2} \end{align}

Answer 1

Note that by standard limits since

$$\frac{a^x-1}{x}\to \log a$$

we have that

$$\frac{3^x+2^x-2}{4^x+2^x-2}=\frac{\frac{3^x-1}x+\frac{2^x-1}x}{\frac{4^x-1}x+\frac{2^x-1}x}\to\frac{\log 3+\log2}{\log 4+\log 2}=\frac{\log 6}{\log 8}$$

Answer 2

\begin{align}\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}&=\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right) \\&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right) \\&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2} \\&= (\ln 3 + \ln 2) \lim_{x\to 0} \frac{x}{4^x+2^x-2} \\&= (\ln 3 + \ln 2) \lim_{x\to 0} \frac{1}{\frac{4^x-1}{x}+\frac{2^x-1}{x}} \\ &=\frac{\ln 3 + \ln 2}{\ln 4 + \ln 2} \\ &= \frac{\ln 6}{\ln 8} \end{align}

Answer 3

$$\lim_{x\rightarrow0}\frac{3^x+2^x-2}{4^x+2^x-2}=\lim_{x\rightarrow0}\frac{\frac{3^x-1}{x}+\frac{2^x-1}{x}}{\frac{4^x-1}{x}+\frac{2^x-1}{x}}=\frac{\ln3+\ln2}{\ln4+\ln2}=\log_86$$ I used $$\lim_{x\rightarrow0}\frac{a^x-1}{x}=\lim_{x\rightarrow0}\left(\frac{e^{x\ln{a}}-1}{x\ln{a}}\cdot\ln{a}\right)=\ln{a}$$ for all $a>0$.

Answer 4

Just added for your curiosity since you already received good answers.

Since $$a^x=e^{x \log(a)}$$ using Taylor series around $x=0$ $$a^x=1+ \log (a)x+\frac{1}{2} \log ^2(a) x^2+O\left(x^3\right)$$ This makes $$A=\frac{3^x+2^x-2}{4^x+2^x-2} =\frac{ (\log (2)+\log (3))x+\frac{1}{2} \left(\log ^2(2)+\log ^2(3)\right)x^2+O\left(x^3\right) } {(\log (2)+\log (4))x+\frac{1}{2} \left(\log ^2(2)+\log ^2(4)\right)x^2+O\left(x^3\right) }$$ Divide top and bottom by $x$ and make the long division to get, after simplifications, $$A=\frac{\log (6)}{\log (8)}-\frac{ \log \left(\frac{4}{3}\right) \log (54)}{18 \log (2)}x+O\left(x^2\right)$$ which shows the limit and how it is approached.

But you can use it for approximation. For illustration, use $x=0.1$. This would give $A\approx 0.852457$ while the exact value would be $A\approx 0.852248$.