I attach the Math Pdf File. Please Open it and Solve my Problem.
Please explain me with step by step.
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The definition of e is
e:=\lim_{n\rightarrow +\infty} \left(1+\frac{1}{n} \right)^n;
this forces that for any sequence of integers \{m_n\}\rightarrow \infty you have
\lim_{n\rightarrow +\infty} \left(1+\frac{1}{m_n} \right)^{m_n}=e

To prove \lim_{x\rightarrow +\infty} \left(1+\frac{1}{x} \right)^x=e,
it is enough to prove that this limit exists. First by direct computation you see that it is a monotone increasing function. Then given any sequence \{x_n\}\rightarrow \infty you have
[x_n]\leq x_n \leq [x_n]+1 thus passing to the limits

e=\lim_{n\rightarrow +\infty} \left(1+\frac{1}{[x_n]} \right)^{[x_n]}\leq
\lim_{n\rightarrow +\infty} \left(1+\frac{1}{x_n} \right)^{x_n}\leq
\lim_{n\rightarrow +\infty} \left(1+\frac{1}{[x_n]+1} \right)^{[x_n]+1}=e.

To prove \lim_{x\rightarrow -\infty} \left(1+\frac{1}{x} \right)^x=e,,
use \lim_{x\rightarrow +\infty} \left(1-\frac{1}{x} \right)^x=\frac{1}{e}
(it comes from the more general characterization \lim_{x\rightarrow +\infty} \left(1+\frac{y}{x} \right)^x=e^y), thus

\lim_{x\rightarrow -\infty} \left(1+\frac{1}{x} \right)^x=
\lim_{x\rightarrow +\infty} \left(1-\frac{1}{x} \right)^{-x}=\left(\frac{1}{e}\right)^{-1}=e.