Answer-0-0-

Question Number 61934 by Cypher1207 last updated on 12/Jun/19

Answer : 0^{0} = ?

Commented by Kunal12588 last updated on 12/Jun/19

I believe 0^{0} = 1

Commented by MJS last updated on 12/Jun/19

x^{x} = 1

x = 1^{\frac{1}{x}}

\lim_{x \to 0^{+}} \frac{1}{x} = + \infty

\lim_{x \to 0^{-}} \frac{1}{x} = - \infty

{\underset{t \to \pm \infty}{\lim 1}}^{t} = 1

\Rightarrow

x = 1

but x = 0 \Rightarrow 0^{0} \neq 1

x^{x} = 1

x \ln x = 0

\Rightarrow x = 0 \lor x = 1

but we could also write

0^{0} = 1

0 \ln 0 = 0

now divide by 0

\ln 0 = \frac{0}{0} = 1 (believing \frac{0}{0} = 1)

\ln 0 = 1

e^{\ln 0} = e^{1}

0 = e

Answered by MJS last updated on 12/Jun/19

0^{0} is not defined

for x \neq 0 : x^{0} = 1 \Rightarrow 0 \ln x = \ln 0 \Rightarrow 0 = 0 true

[\ln x \in C for x < 0 \Rightarrow 0 \ln x = 0]

for x > 0 : 0^{x} = 0 \Rightarrow 0 = \sqrt[x]{0} \Rightarrow 0 = 0 true

[for x < 0 we get {(\frac{1}{0})}^{x} which {isn}^{'} t defined too]

\lim_{x \to 0} x^{0} = 1

\lim_{x \to 0^{+}} 0^{x} = 0

Commented by MJS last updated on 12/Jun/19

but working on a special function it can be

useful to define it

y = x^{0} \Rightarrow 0^{0} := 1

y = 0^{x} \Rightarrow 0^{0} := 0

y = x^{x} \Rightarrow 0^{0} := 1

similar problems occur with \frac{0}{0}

y = \frac{0}{x} \Rightarrow \frac{0}{0} := 0

y = \frac{x}{x} \Rightarrow \frac{0}{0} := 1

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