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Question Number 125292 by Kurbanklichevs last updated on 09/Dec/20

lim_(x→0)  (((x+1)^(n+1) −(n+1)(x+1)+n)/x^2 )     n∈N  Calculate. (Without a L′Hopital′s rule)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{n}}{{x}^{\mathrm{2}} }\:\:\:\:\:{n}\in\mathbb{N} \\ $$$${Calculate}.\:\left({Without}\:{a}\:{L}'{Hopital}'{s}\:{rule}\right) \\ $$

Answered by Dwaipayan Shikari last updated on 09/Dec/20

lim_(x→0) (((1+x)^(n+1) −(n+1)(x+1)+n)/x^2 )  lim_(x→0) ((1+(n+1)x+((n(n+1))/(2!))x^2 −(n+1)(x+1)+n)/x^2 )  lim_(x→0) ((1−n−1−n+((n(n+1))/(2!))x^2 )/x^2 )  =((n(n+1))/2)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{n}}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\left({n}+\mathrm{1}\right){x}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{n}}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{n}−\mathrm{1}−{n}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$

Commented by Kurbanklichevs last updated on 10/Dec/20

Tank you

$${Tank}\:{you} \\ $$

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