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Maths-

General

Easy

Question

$L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus e to the power of s in space x end exponent over denominator 2 left parenthesis x minus sin space x right parenthesis end fraction equals$

1
2
-1
$\frac{1}{2}$

hint Hint:

In this question we have the value for this we will simplify the given equation and then put the value of x.

The correct answer is: $1 half$

$L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus e to the power of s in space x end exponent over denominator 2 left parenthesis x minus sin space x right parenthesis end fraction equals L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of sin x end exponent open parentheses e to the power of x minus sin x end exponent minus 1 close parentheses over denominator 2 open parentheses x minus sin space x close parentheses end fraction equals L t subscript x not stretchy rightwards arrow 0 end subscript e to the power of sin space x end exponent over 2 equals e to the power of sin space 0 end exponent over 2 equals e to the power of 0 over 2 equals 1 half.$

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