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lim-x-0-1-x-a-1-x-for-a-R-Don-t-using-L-hospital-rules-

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Question Number 25379 by A1B1C1D1 last updated on 09/Dec/17
lim_(x → 0) ((((1 + x)^a − 1)/x)) for a ∈ R Don′t using L′hospital rules.
$$\underset{\mathrm{x}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{1}\:+\:\mathrm{x}\right)^{\mathrm{a}} \:−\:\mathrm{1}}{\mathrm{x}}\right)\:\mathrm{for}\:\mathrm{a}\:\in\:\mathbb{R} \\ $$$$ \\ $$$$\mathrm{Don}'\mathrm{t}\:\mathrm{using}\:\mathrm{L}'\mathrm{hospital}\:\mathrm{rules}. \\ $$
Answered by Rasheed.Sindhi last updated on 10/Dec/17
lim_(x→0) ((((1 + x)^n − 1)/x)) lim_(x→0) ((((1 +(n/1) x+((n(n−1))/(1.2))x^2 +((n(n−1)(n−2))/(1.2.3))x^3 +...− 1)/x)) lim_(x→0) (((((n/1) x+((n(n−1))/(1.2))x^2 +((n(n−1)(n−2))/(1.2.3))x^3 +...))/x)) lim_(x→0) ((n/1) +((n(n−1))/(1.2))x+((n(n−1)(n−2))/(1.2.3))x^2 +...) =n
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{1}\:+\:\mathrm{x}\right)^{\mathrm{n}} \:−\:\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{1}\:+\frac{\mathrm{n}}{\mathrm{1}}\:\mathrm{x}+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\mathrm{x}^{\mathrm{3}} +…−\:\mathrm{1}\right.}{\mathrm{x}}\right) \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\left(\frac{\mathrm{n}}{\mathrm{1}}\:\mathrm{x}+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\mathrm{x}^{\mathrm{3}} +…\right)}{\mathrm{x}}\right) \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{n}}{\mathrm{1}}\:+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{2}}\mathrm{x}+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\mathrm{x}^{\mathrm{2}} +…\right) \\ $$$$=\mathrm{n} \\ $$
Commented by A1B1C1D1 last updated on 09/Dec/17
Thank you
$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by moxhix last updated on 10/Dec/17
another way put f(x)=(1+x)^a f ′(x)=a(1+x)^(a−1) →f ′(0)=a f ′(0)=lim_(x→0) (((1+x)^a −(1+0)^a )/(x−0)) ∴lim_(x→0) (((1+x)^a −1)/x)=a
$${another}\:{way} \\ $$$${put}\:{f}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{a}} \\ $$$${f}\:'\left({x}\right)={a}\left(\mathrm{1}+{x}\right)^{{a}−\mathrm{1}} \rightarrow{f}\:'\left(\mathrm{0}\right)={a} \\ $$$${f}\:'\left(\mathrm{0}\right)=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{x}\right)^{{a}} −\left(\mathrm{1}+\mathrm{0}\right)^{{a}} }{{x}−\mathrm{0}} \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{x}\right)^{{a}} −\mathrm{1}}{{x}}={a} \\ $$

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