Eagerly waiting so plzz reply fast ..............
in limits if e^n -1/n=1
and we know that e has the value 2.7
so it is the constant does'nt it???
so why not every constant has value 1 like
1^n-1/n=1 why it is not possible???
any queries about the question yeah you can ask me in comment section given below
but ya i want an answer.............

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Answers

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2014-11-14T01:02:10+05:30

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Euler's number e is defined as follows:

e= \lim_{m \to \infty} (1+\frac{1}{m})^m,\ \ m=\frac{1}{y},\ \ y-> 0\\\\= \lim_{y \to 0} (1+y)^m\\\\for\ example\ e = approximately (1+0.001)^{1000}\\\\\frac{e^x-1}{x} =\lim_{y \to 0} \frac{(1+y)^{mx}-1 }{x}\\

Let n = mx = x/y. Use binomial expansion theorem,

=\lim_{y \to 0}, \frac{(1+ {}^nC_1\ x+ {}^nC_2\ x^2+....+{}^nC_n\ x^n)-1}{x}\\\\=\lim_{x \to 0},\ \lim_{n \to \infty} \frac{{}^nC_1\ x+ {}^nC_2\ x^2+....+{}^nC_n\ x^n}{x}\\\\=\lim_{x \to 0},\ \lim_{n \to \infty} {}^nC_1+ {}^nC_2\ x+....+{}^nC_n\ x^{n-1}\\\\={}^nC_1=n=x/y\\

 \lim_{x \to 0}, \lim_{y \to 0} \frac{e^x-1}{x} = n=\frac{x}{y}\\\\while\ choosing\ limits,\ we\ choose\ x=y,\ as\ both\ near\ 0.\\\\\lim_{x \to 0} \frac{e^x-1}{x}=1\\

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Let a = a positive real number,  like 1, e , 2

Let\ z = a^x\\\\Log_e\ z = x\ (Log_e\ a)\\\\So,\ z=a^x=e^{(x\ Log_e\ a)}= (e^x )^{Log_e\ a}\\\\ \lim_{x \to 0} \frac{a^x -1}{x} =\frac{(e^x )^{Log_e\ a}-1}{x}\\\\The\ procedure\ is\ same\ as\ above,\ with\ n=(m*x*Log_e\ a)\\\\Or,\ n=\frac{x}{y}*Log_e\ a\\\\Hence, \lim_{x \to 0} \frac{a^x -1}{x} = n=\frac{x}{y}*Log_e\ a=Log_e\ a\\\\

Now, choose a = 1,  then, you get :
\lim_{x \to 0} \frac{a^x -1}{x}=Log_e\ 1=0\\

If you choose a = e, then you get

\lim_{x \to 0} \frac{e^x -1}{x}=Log_e\ e=1\\\\If\ a=10,\ then\\\\\lim_{x \to 0} \frac{10^x -1}{x}=Log_e\ 10\\
1 5 1
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e^x is also defined using Taylor series as: e^x=1+x/1!+x^2/2!+....to infinity. This is not finite series. e^x-1 = x(1+x/2! +x^2/3!+..... to infinity. (e^x-1)/x = 1+x/2!+x^2/3!+....infinity, as x ->0, RHS = 1.