The correct option is B 2 lim_x→ 0e^x 1√(1+x) 1 Using L hospital's rule, we get lim_x→ 0e^x12√(1+x)=lim_x→ 0112=2 ...

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Laws of Exponents for Real Numbers

limx→ 0ex-1/√...

Question

$lim x \to 0 \frac{e^{x} - 1}{\sqrt{1 + x} - 1} =$

1

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\frac{1}{2}

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\frac{1}{3}

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2

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Solution

The correct option is B $2$
$lim x \to 0 \frac{e^{x} - 1}{\sqrt{1 + x} - 1}$
Using L-hospital's rule, we get
$lim x \to 0 \frac{e^{x}}{\frac{1}{2 \sqrt{1 + x}}} = lim x \to 0 \frac{1}{\frac{1}{2}} = 2$

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