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I need to find $\lim_{h \to 0, h\ne 0} \sqrt[h]{\frac{3^h+2^h}{2}}$.

My attempt:

$\lim_{h \to 0, h\ne 0} \sqrt[h]{\frac{3^h+2^h}{2}}=\lim_{h \to 0, h\ne 0}\exp(\frac{\log(\frac{2^h+3^h}{2})}{h})=\exp(\lim_{h \to 0, h\ne 0}\frac{\log(\frac{2^h+3^h}{2})}{h})$

$\lim_{h \to 0, h\ne 0}\frac{\log(\frac{2^h+3^h}{2})}{h}= \lim_{h \to 0, h\ne 0}\frac{\log(\frac{2^h+3^h}{2})-\log(\frac{2^0+3^0}{2})}{h-0}=(\log(\frac{2^h+3^h}{2}))'(0)=\frac{1}{2^0+3^0}0(2^{(0-1)}+3^{(0-1)})=0$

So $\lim_{h \to 0, h\ne 0} \sqrt[h]{\frac{3^h+2^h}{2}}=\exp(0)=1$

Was my solution correct? I am asking because I tried to check numerically with Matlab and have not noticed the convergence to $1$.

zesy
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    You do not have to add $h\neq0$. The limit already does that. – Arbuja Jan 24 '16 at 13:54
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    Two observations: i) you computed the derivative incorrectly, as $\frac{d}{dx}a^x=a^x\ln a$ (see my answer below), and ii) putting $h\neq 0$ as a subscript is bad notation. – Pierpaolo Vivo Jan 24 '16 at 13:55

2 Answers2

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$$ \frac{d}{dx}\ln(\frac{2^x+3^x}{2})=\frac{1}{2^x+3^x}(2^x\ln2+3^x\ln 3) $$ which for $x=0$ gives $(\ln2+\ln 3)/2$. Therefore your limit is $\sqrt{6}$.

Pierpaolo Vivo
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$$\lim_{h\to0}\sqrt[h]{\frac{3^h+2^h}{2}}=$$ $$\lim_{h\to0}\left(\frac{3^h+2^h}{2}\right)^{\frac{1}{h}}=$$ $$\lim_{h\to0}\exp\left[\ln\left(\left(\frac{3^h+2^h}{2}\right)^{\frac{1}{h}}\right)\right]=$$ $$\lim_{h\to0}\exp\left[\frac{1}{h}\ln\left(\frac{3^h+2^h}{2}\right)\right]=$$ $$\exp\left[\lim_{h\to0}\frac{\ln\left(\frac{3^h+2^h}{2}\right)}{h}\right]=$$ $$\exp\left[\lim_{h\to0}\frac{\frac{\text{d}}{\text{d}h}\left(\ln\left(\frac{3^h+2^h}{2}\right)\right)}{\frac{\text{d}}{\text{d}h}\left(h\right)}\right]=$$ $$\exp\left[\lim_{h\to0}\frac{\frac{2^h\ln(2)+3^h\ln(3)}{2^h+3^h}}{1}\right]=$$ $$\exp\left[\frac{\frac{2^0\ln(2)+3^0\ln(3)}{2^0+3^0}}{1}\right]=$$ $$\exp\left[\frac{\frac{\ln(2)+\ln(3)}{1+1}}{1}\right]=$$ $$\exp\left[\frac{\frac{\ln(2)+\ln(3)}{2}}{1}\right]=\exp\left[\frac{\ln(2)+\ln(3)}{2}\right]=\exp\left[\frac{\ln(6)}{2}\right]=\sqrt{6}$$

Jan Eerland
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