Question Number 26591 by prakash jain last updated on 27/Dec/17
$$\mathrm{Prove} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\mathrm{0} \\ $$
Commented by abdo imad last updated on 27/Dec/17
$${let}\:{put}\:\mathrm{1}+{x}={u}\:\:\:\Rightarrow\:{lim}_{{x}−>−\mathrm{1}^{+} } \:\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)\:\:={lim}_{{u}−>\mathrm{0}^{+} } \:\:{ulnu}\:=\mathrm{0} \\ $$
Answered by ajfour last updated on 27/Dec/17
![L=lim_(x→−1) ((ln (1+x))/(((1/(1+x))))) =lim_(x→−1) [−(((1+x)^2 )/(1+x))]=−lim_(x→−1) (1+x)=0 .](Q26598.png)
$${L}=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)} \\ $$$$\:=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\left[−\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{\mathrm{1}+{x}}\right]=−\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\left(\mathrm{1}+{x}\right)=\mathrm{0}\:. \\ $$