- #1
coldboyqn
- 8
- 0
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 anyone 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 anyone show me, please?
Last edited: