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Question-71096




Question Number 71096 by Omer Alattas last updated on 11/Oct/19
Answered by $@ty@m123 last updated on 11/Oct/19
Let x=1+y  lim_(y→0)     (y/(ln (1+y)))    (1/(lim_(y→0)  ((ln (1+y))/y)))=1
$${Let}\:{x}=\mathrm{1}+{y} \\ $$$$\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\:\frac{{y}}{\mathrm{ln}\:\left(\mathrm{1}+{y}\right)} \\ $$$$\:\:\frac{\mathrm{1}}{\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+{y}\right)}{{y}}}=\mathrm{1} \\ $$
Commented by Omer Alattas last updated on 11/Oct/19
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Commented by $@ty@m123 last updated on 11/Oct/19
lim_(x→0)    ((ln (1+x))/x)=1 is a standard limit formula.  It can be proved easily using  logarithmic expansion.  i.e. ln (1+x)=x−(x^2 /2)+(x^3 /3)− ....
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}=\mathrm{1}\:{is}\:{a}\:{standard}\:{limit}\:{formula}. \\ $$$${It}\:{can}\:{be}\:{proved}\:{easily}\:{using} \\ $$$${logarithmic}\:{expansion}. \\ $$$${i}.{e}.\:\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\:…. \\ $$
Commented by malwaan last updated on 11/Oct/19
lim_(x→0) ((ln(1+x))/x)  = lim_(x→0) ((x−(x^2 /2)+(x^3 /3)−(x^4 /4)+...)/x)  = lim_(x→0) ((1−(x/2)+(x^2 /3)−(x^3 /4)+...)/1)  = 1−0+0−0+...=1
$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}{\boldsymbol{{x}}} \\ $$$$=\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\boldsymbol{{x}}−\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{2}}+\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{3}}−\frac{\boldsymbol{{x}}^{\mathrm{4}} }{\mathrm{4}}+…}{\boldsymbol{{x}}} \\ $$$$=\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\mathrm{1}−\frac{\boldsymbol{{x}}}{\mathrm{2}}+\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{3}}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{4}}+…}{\mathrm{1}} \\ $$$$=\:\mathrm{1}−\mathrm{0}+\mathrm{0}−\mathrm{0}+…=\mathrm{1} \\ $$

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