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Maths-

General

Easy

Question

$Lt subscript x not stretchy rightwards arrow 0 end subscript invisible function application fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction$

1
0
2
$\frac{1}{2}$

hint Hint:

In this question firstly we will rationalize and simplify the equation. later we will put the value of x to find the required answer.

The correct answer is: 2

$space space space space Lt subscript x not stretchy rightwards arrow 0 end subscript invisible function application fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction equals space Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction cross times fraction numerator square root of 1 plus x end root plus 1 over denominator square root of 1 plus x end root plus 1 end fraction equals Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 open parentheses square root of 1 plus x end root plus 1 close parentheses over denominator 1 plus x minus 1 end fraction equals Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 open parentheses square root of 1 plus x end root plus 1 close parentheses over denominator x end fraction equals Lt subscript x not stretchy rightwards arrow 0 end subscript space open parentheses square root of 1 plus x end root plus 1 close parentheses equals square root of 1 plus 0 end root plus 1 equals 1 plus 1 equals 2.$

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