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I am doing something that I sure that I'm wrong, but I cannot realize the error. See as below:

[tex]

\frac{1}{x}=x\times\frac{1}{(x^2)} ________\(1\)

[/tex]

Taylor Series of [tex]\frac{1}{x^2}[/tex]:

[tex]

\frac{1}{x^2}=\frac{1}{\alpha}+\sum_{k=1}^\infty g(k)(x-\alpha)^k

[/tex]

In which

k is from 1 to infinity,

[tex]

g(k)=(-1)^k\times\frac{(k+1)!}{\alpha^{(k+2)}k!}\\\\

=\frac{(-1)^k\times(k+1)}{\alpha^{(k+2)}}

[/tex]

Substitute Taylor Series of 1/x^2 into (1), we obtain:

[tex]\frac{1}{x}=\frac{x}{\alpha}+\sum_{k=1}^\infty g(k)(x-\alpha)^k[/tex]

So: [tex]\lim_{\substack{x\rightarrow 0}} \frac{1}{x}=\lim_{\substack{x\rightarrow 0}} (\frac{x}{\alpha}+\sum_{k=1}^\infty x\times g(k)(x-\alpha)^k)=0 (!?!?!?)

[/tex]

Can any one show me, please?

[tex]

\frac{1}{x}=x\times\frac{1}{(x^2)} ________\(1\)

[/tex]

Taylor Series of [tex]\frac{1}{x^2}[/tex]:

[tex]

\frac{1}{x^2}=\frac{1}{\alpha}+\sum_{k=1}^\infty g(k)(x-\alpha)^k

[/tex]

In which

k is from 1 to infinity,

[tex]

g(k)=(-1)^k\times\frac{(k+1)!}{\alpha^{(k+2)}k!}\\\\

=\frac{(-1)^k\times(k+1)}{\alpha^{(k+2)}}

[/tex]

Substitute Taylor Series of 1/x^2 into (1), we obtain:

[tex]\frac{1}{x}=\frac{x}{\alpha}+\sum_{k=1}^\infty g(k)(x-\alpha)^k[/tex]

So: [tex]\lim_{\substack{x\rightarrow 0}} \frac{1}{x}=\lim_{\substack{x\rightarrow 0}} (\frac{x}{\alpha}+\sum_{k=1}^\infty x\times g(k)(x-\alpha)^k)=0 (!?!?!?)

[/tex]

Can any one show me, please?

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