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Question-148819




Question Number 148819 by mathlove last updated on 31/Jul/21
Answered by liberty last updated on 31/Jul/21
 lim_(x→0)  (x/((1+(x^2 /n))−(1+(x/n)))) =   lim_(x→0)  (x/((x/n)(x−1))) = lim_(x→0)  (n/(x−1))=− n
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}}{\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}}\right)−\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{n}}\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}}{\frac{\mathrm{x}}{\mathrm{n}}\left(\mathrm{x}−\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{n}}{\mathrm{x}−\mathrm{1}}=−\:\mathrm{n} \\ $$
Answered by Kamel last updated on 31/Jul/21
L=lim_(x→0) (1/( ((((1+x^2 ))^(1/n) −1)/x)−((((1+x))^(1/n) −1)/x)))=(1/(−(1/n)))=−n
$${L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{\:\frac{\sqrt[{{n}}]{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}}{{x}}−\frac{\sqrt[{{n}}]{\mathrm{1}+{x}}−\mathrm{1}}{{x}}}=\frac{\mathrm{1}}{−\frac{\mathrm{1}}{{n}}}=−{n} \\ $$

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