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Question Number 116037 by Rio Michael last updated on 30/Sep/20

$$\underset{x\to 0}{\lim }\frac{{27}^x-1}{9^x-1}=??$$

Answered by bemath last updated on 30/Sep/20

$$let{3}^x=t;t\to 1$$$$\underset{t\to 1}{\lim }\frac{t^3-1}{t^2-1}=\underset{t\to 1}{\lim }\frac{(t-1)(t^2+t+1)}{(t-1)(t+1)}=\frac{3}{2}$$

Commented by Rio Michael last updated on 30/Sep/20

$$\mathrm{exactly}.$$$$\underset{x\to 0}{\lim }\frac{{27}^x-1}{9^x-1}=\underset{x\to 0}{\lim }\frac{(3^x-1)(3^{2x}+\left(3^x\right)+1)}{(3^x-1)(3^x+1)}=\frac{3}{2}$$$$$$

Commented by bemath last updated on 30/Sep/20

$$yes..$$

Answered by Dwaipayan Shikari last updated on 30/Sep/20

$$\underset{x\to 0}{\lim }\frac{{27}^{\mathrm{x}}-1}{\mathrm{x}}.\frac{\mathrm{x}}{9^{\mathrm{x}}-1}=\frac{\log \left(27\right)}{\log \left(9\right)}=\frac{3}{2}$$

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