Which of the following limit is equal to 0?

$lim x \to 0^{+} (x^{x^{x}} - x^{x})$

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$lim x \to 0^{+} x^{2} ln (\sqrt{\frac{1}{x}})$

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$lim x \to 0^{+} (x)^{ln (x + 1)}$

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$lim x \to 0 (\frac{10^{x} - 2^{x} - 5^{x} + 1^{x}}{x + tan x})$

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Solution

The correct options are
B
$lim x \to 0^{+} x^{2} ln (\sqrt{\frac{1}{x}})$

D
$lim x \to 0 (\frac{10^{x} - 2^{x} - 5^{x} + 1^{x}}{x + tan x})$

$y = lim x \to 0^{+} x^{x} \Rightarrow y = lim x \to 0^{+} e^{x ln x} = e^{lim x \to 0^{+} x ln x} = e^{lim x \to 0^{+} \frac{ln x}{\frac{1}{x}}} = e^{lim x \to 0^{+} \frac{\frac{1}{x}}{- \frac{1}{x^{2}}}} = e^{lim x \to 0^{+} - x} = e^{lim x \to 0^{+} 0} = 1$

$∴ lim x \to 0^{+} (x^{x^{x}} - x^{x}) = lim x \to 0^{+} x^{x^{x}} - lim x \to 0^{+} x^{x} = e^{lim x \to 0^{+} x^{x} ln x} - 1 = e^{- \infty} - 1 = 0 - 1 = - 1$

$lim x \to 0^{+} x^{2} ln (\sqrt{\frac{1}{x}}) = (- \frac{1}{2}) lim x \to 0^{+} x^{2} ln x . . . . . [0 \times \infty f o r m]$

$= (- \frac{1}{2}) lim x \to 0^{+} \frac{ln x}{(\frac{1}{x^{2}})} . . . . (\frac{\infty}{\infty}$ form)

$= (- \frac{1}{2}) lim x \to 0^{+} \frac{(\frac{1}{x})}{(\frac{- 2}{x^{3}})} = \frac{1}{4} lim x \to 0^{+} (x^{2}) = 0$

Let $ℓ = lim x \to 0^{+} (x)^{ln (x + 1)} \Rightarrow ln ℓ = lim x \to 0^{+} \frac{ln x}{[ln (x + 1)]^{- 1}}$

$= lim x \to 0^{+} \frac{(\frac{1}{x})}{(\frac{- 1}{{ln}^{2} (x + 1)}) (\frac{1}{x + 1})} = lim x \to 0^{+} \frac{(x + 1) {ln}^{2} (x + 1)}{- x}$

$= - lim x \to 0^{+} (\frac{ln (x + 1)}{x})^{2} (x^{2} + x) = - (1)^{2} (0 + 0) = 0$

Hence, $ln ℓ = 0$

$\Rightarrow ℓ = 1$

$lim x \to 0 \frac{10^{x} - 2^{x} - 5^{x} + 1^{x}}{x + tan x} = lim x \to 0 \frac{x (\frac{5^{x} - 1}{x}) (\frac{2^{x} - 1}{x})}{(1 + \frac{t a n x}{x})} = \frac{(0) (1) (1)}{2} = 0$

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