Asked by Topperlearning User | 28th Jul, 2014, 10:34: AM

Expert Answer:

limit as blank rightwards arrow 0 of open parentheses fraction numerator sin space x over denominator x end fraction close parentheses to the power of 1 over x squared end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator log open parentheses begin display style fraction numerator sin space x over denominator x end fraction end style close parentheses over denominator x squared end fraction space space open square brackets 0 over 0 f o r m close square brackets end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator begin display style fraction numerator x over denominator sin space x end fraction end style. begin display style fraction numerator x. cos space x minus sin space x over denominator x squared end fraction end style over denominator 2 x end fraction space open square brackets L apostrophe H o s p i t a l apostrophe s space R u l e close square brackets end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator x. cos space x minus sin space x over denominator 2 x squared sin space x end fraction open square brackets 0 over 0 space f o r m close square brackets end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator minus x. sin space x plus cos space x minus cos space x over denominator 4 x. sin space x plus 2 x squared. cos space x end fraction end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator minus x. sin space x over denominator 2 x. left parenthesis 2 sin space x plus x. cos space x right parenthesis end fraction end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator minus sin space x over denominator 2 left parenthesis 2 sin space x plus x. cos space x right parenthesis end fraction end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator minus sin space x over denominator 2 left parenthesis 2 sin space x plus x. cos space x right parenthesis end fraction open square brackets 0 over 0 space f o r m close square brackets end exponent
equals e to the power of limit as x rightwards arrow 0 of fraction numerator minus cos space x over denominator 2 left parenthesis 2 cos space x plus cos space x minus x. sin space x right parenthesis end fraction end exponent
equals e to the power of fraction numerator minus 1 over denominator 2 left parenthesis 3 minus 0 right parenthesis end fraction end exponent equals e to the power of minus 1 over 6 end exponent

Answered by  | 28th Jul, 2014, 12:34: PM

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