Question

Well, I feel I just probably might have discovered something, please check it below...
$y \: = \: {x}^{y}$
$log_{x}(y) \: = \: y$
$\frac{y}{ log_{x}(y) } \: = \: 1$
$y \: \times \: log_{y}(x) \: = \: 1$
Differentiating w.r.t. x on both sides, we get,
$\frac{d(y \times log_{y}(x) )}{dx} \: = \: \frac{d(1)}{dx}$
$y \: \times \: \frac{d( log_{y}(x) )}{dx} \: = \: 0$
$y \: \times \: \frac{1}{x \: \times \: ln(y) } \: = \: 0$
From the above equation, we can conclude that,
$y \: = \: 0$
$\frac{1}{x} \: = \: 0$
$\frac{1}{ ln(y) } \: = \: 0$
or,
$ln(y) \: = \: \frac{1}{0}$
Therefore, by taking Anti-Logarithm,
${e}^{ \frac{1}{0} } \: = \: y$
Since,
$y \: = \: 0$
Hence,
${e}^{ \frac{1}{0} } \: = \: 0$
or,
${e}^{ \frac{1 \: \times \: 0}{0}} \: = \: {0}^{0}$
${0}^{0} \: = \: {0}^{x - x} \: = \: \frac{ {0}^{x} }{ {0}^{x} } \: = \: \frac{0}{0}$
or,
${0}^{0} \: = \: \frac{0}{0}$
Following Leonhard Euler's Proof below,
$\frac{lim}{x \: - > \: 0} \: {x}^{x} \: = \: 1$
${e}^{1} \: = \: 1$
or,
$e \: = 1 ?$
Wait... What??? What did I just prove???

Answer 1

Pls ask proper questions.

Thnks!

Shauna

Answer 2

nice dear..
congerats
ye talents bhut kam m hote h..
keep it up
be brainly...
be genius...
But u are requested to ask proper questions