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Question Number 106435 by Study last updated on 05/Aug/20

$$li\underset{x\to 2}{m}\frac{ln(2x+10)}{x-2}=??$$

Commented by Dwaipayan Shikari last updated on 05/Aug/20

$$limit{doesn}^{\prime }texist$$

Commented by bemath last updated on 05/Aug/20

$$\ln \left(14\right)\ne 0.\mathrm{so}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{rule}{\mathrm{doesn}}^{\prime }\mathrm{t}$$$$\mathrm{work}.$$

Answered by 1549442205PVT last updated on 05/Aug/20

$$\mathrm{Set}\mathrm{x}-2=\mathrm{t}\mathrm{we}\mathrm{get}li\underset{x\to 2}{m}\frac{ln(2x+10)}{x-2}=$$$$\underset{\mathrm{x}\to 2}{\lim }\frac{\ln [2(\mathrm{x}-2)+14]}{\mathrm{x}-2}=\underset{\mathrm{t}\to 0}{\lim }\frac{\ln (2\mathrm{t}+14)}{\mathrm{t}}$$$$\underset{\mathrm{Hopital}}{=}\underset{\mathrm{t}\to 0}{\mathbf{lim}}\frac{2}{2\mathbf{t}+14}=\frac{2}{14}=\frac{1}{7}$$

Commented by Her_Majesty last updated on 05/Aug/20

$$youcannotuse{l}^{\prime }Hopitalinthiscasebecause$$$$ln(2t+14)isdefinedfort=0$$

Commented by 1549442205PVT last updated on 05/Aug/20

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}.\mathrm{I}\mathrm{mistaked}.\mathrm{That}\mathrm{limit}$$$$\mathrm{lead}\mathrm{to}\mathrm{\infty }$$

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