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Question Number 182836 by mathlove last updated on 15/Dec/22

Answered by floor(10²Eta[1]) last updated on 15/Dec/22

$$\underset{x\to 2}{\lim }\frac{2^{\mathrm{x}}\ln 2+3^{\mathrm{x}}\ln 3+4^{\mathrm{x}}\ln 4}{1+2\mathrm{x}+{3\mathrm{x}}^2+{4\mathrm{x}}^3}=\frac{4\ln 2+9\ln 3+16\ln 4}{1+4+12+32}$$$$=\frac{36\ln 2+9\ln 3}{49}$$

Commented by mathlove last updated on 17/Dec/22

$$without{H}^{\prime }pitalRoul$$

Answered by manolex last updated on 15/Dec/22

$$evaluatex=2h\left(2\right)=\frac{0}{0}$$$${lim}_{x\to 2}h\left(x\right)=\frac{0}{0}\Rightarrow aplique{L}^{\prime }hospital$$$$\frac{f^{\prime }}{g^{\prime }}=\frac{0+\left(ln2\right)2^x+\left(ln3\right)3^x+\left(ln4\right)4^x}{1+2x+3x^2+4x^3}$$$${lim}_{x\to 2}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\frac{4ln2+9ln3+16ln4}{1+4+12+32}=\frac{36ln\left(2\right)+9ln3}{49}=0.71103$$$$evaluatewhitcalculator$$$$h{(2.01)}_{+}=0.711098$$$$h{(1.99)}_{-}=0.711099$$$$ok,right$$$$$$$$$$

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