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Maths-

General

Easy

Question

$L t subscript x not stretchy rightwards arrow 1 end subscript fraction numerator log subscript e invisible function application open parentheses 1 plus x cubed close parentheses over denominator sin cubed invisible function application x end fraction$

logx
1
x³
0

The correct answer is: 1

$limit as x rightwards arrow 1 of fraction numerator log open parentheses 1 plus x cubed close parentheses over denominator sin cubed x end fraction W e space k n o w space t h a t comma limit as x rightwards arrow 1 of fraction numerator log open parentheses 1 plus x close parentheses over denominator x end fraction equals 1 space a n d space limit as x rightwards arrow 1 of fraction numerator sin x over denominator x end fraction equals 1 T h e r e f o r e comma limit as x rightwards arrow 1 of fraction numerator log open parentheses 1 plus x cubed close parentheses over denominator sin cubed x end fraction D i v i d i n g space n u m e r a t o r space a n d space d e n o m i n a t o r space b y space x cubed comma equals limit as x rightwards arrow 1 of fraction numerator begin display style fraction numerator log open parentheses 1 plus x cubed close parentheses over denominator x cubed end fraction end style over denominator begin display style fraction numerator sin cubed x over denominator x cubed end fraction end style end fraction equals fraction numerator begin display style limit as x rightwards arrow 1 of open parentheses fraction numerator log open parentheses 1 plus x cubed close parentheses over denominator x cubed end fraction close parentheses end style over denominator begin display style limit as x rightwards arrow 1 of open parentheses fraction numerator sin cubed x over denominator x cubed end fraction close parentheses end style end fraction equals 1 over 1 equals 1$

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