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Question Number 199908 by cortano12 last updated on 11/Nov/23

$$\underset{x\to 0}{\lim }{\left(\frac{2^{\mathrm{x}}+3^{\mathrm{x}}+5^{\mathrm{x}}}{3}\right)}^{\frac{3}{\mathrm{x}}}=?$$

Answered by cortano12 last updated on 11/Nov/23

$$\mathrm{L}=\underset{x\to 0}{\lim }{\left(\frac{2^{\mathrm{x}}+3^{\mathrm{x}}+5^{\mathrm{x}}}{3}\right)}^{\frac{3}{\mathrm{x}}}$$$$\ln \mathrm{L}=\underset{x\to 0}{\lim }\frac{3}{\mathrm{x}}(\ln (2^{\mathrm{x}}+3^{\mathrm{x}}+5^{\mathrm{x}})-\ln 3)$$$$\ln \mathrm{L}=3.\underset{x\to 0}{\lim }\frac{\ln (2^{\mathrm{x}}+3^{\mathrm{x}}+5^{\mathrm{x}})-\ln 3}{\mathrm{x}}$$$$\ln \mathrm{L}=3.\frac{\mathrm{d}}{\mathrm{dx}}\ln (2^{\mathrm{x}}+3^{\mathrm{x}}+5^{\mathrm{x}}){\mid }_{\mathrm{x}=0}$$$$\ln \mathrm{L}=3.\mid \frac{2^{\mathrm{x}}\ln 2+3^{\mathrm{x}}\ln 3+5^{\mathrm{x}}\ln 5}{2^{\mathrm{x}}+3^{\mathrm{x}}+5^{\mathrm{x}}}{\mid }_{\mathrm{x}=0}$$$$\ln \mathrm{L}=\ln 2+\ln 3+\ln 5\Rightarrow \mathrm{L}=30$$

Answered by MM42 last updated on 11/Nov/23

$$=e^{{lim}_{x\to 0}(\frac{2^x+3^x+5^x}{3}-1)\times \frac{3}{x}}$$$$=e^{\stackrel{hop}{\to }{lim}_{x\to 0}\left(\frac{2^{x}ln2+3^{x}ln3+5^{x}ln5}{1}\right)}$$$$=e^{ln30}=30✓$$

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