x→ 0 ^ + lim x ^ x = 0 ^ 0 = e ^ln x ^ x #160; = e ^ xln x #160; = e ^ 0( ∞ #160; ) #160; blog _ b X =x log _ b X ^ N =Nlog _ b ...

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Substitution Method to Remove Indeterminate Form

Find : limx ...

Question

Find :
$lim x \to 0^{+} x^{x}$

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Solution

$l i m x \to 0^{+} x^{x} = 0^{0} = e^{ln x^{x}} = e^{x ln x} = e^{0 (- \infty)} b {log}_{b} X = x {log}_{b} X^{N} = N {log}_{b} x ∴ l i m x \to 0^{+} x^{x} = e^{\frac{ln x}{\frac{1}{x}}} = e^{- \frac{\infty}{\infty}} = e^{\frac{\frac{1}{x}}{- \frac{1}{x^{2}}}} = e^{\frac{1}{x} \cdot - \frac{x^{2}}{1}} = e^{- x} = e^{0} = 1$

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