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Maths-

General

Easy

Question

$L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator open parentheses 1 minus e to the power of x close parentheses sin begin display style space end style x over denominator x squared plus x cubed end fraction equals$

1
0
-1
$e^{2}$

hint Hint:

In this question, we have to the find the value of $L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator open parentheses 1 minus e to the power of x close parentheses sin begin display style space end style x over denominator x squared plus x cubed end fraction$ . Firstly, we will split the limit into different easy solvable way. Then using different formulas $open square brackets space i e. comma space space space limit as x rightwards arrow 0 of open parentheses fraction numerator e to the power of x minus 1 over denominator x end fraction close parentheses equals 1 comma space space limit as x rightwards arrow 0 of open parentheses fraction numerator sin space x over denominator x end fraction close parentheses equals 1 comma space space space space limit as x rightwards arrow 0 of open parentheses fraction numerator 1 over denominator 1 plus x end fraction close parentheses space close square brackets$ we can find the required value.

The correct answer is: -1

$limit as x rightwards arrow 0 of fraction numerator open parentheses 1 minus e to the power of x close parentheses sin space x over denominator x squared plus x cubed end fraction equals limit as x rightwards arrow 0 of fraction numerator negative open parentheses e to the power of x minus 1 close parentheses sin space x over denominator x squared left parenthesis 1 plus x right parenthesis end fraction equals space space space minus space space space limit as x rightwards arrow 0 of open parentheses fraction numerator e to the power of x minus 1 over denominator x end fraction close parentheses. limit as x rightwards arrow 0 of open parentheses fraction numerator sin space x over denominator x end fraction close parentheses. limit as x rightwards arrow 0 of open parentheses fraction numerator 1 over denominator 1 plus x end fraction close parentheses equals negative open square brackets 1.1. fraction numerator 1 over denominator 1 plus 0 end fraction close square brackets equals negative 1$

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