Math, asked by pritamsarbajna10, 4 months ago

what is the answer ??​

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Answered by Anonymous
31

 \binom{lim}{x →0  } \frac{x \cos(x) +  \sin(x)  }{ {x}^{2}  +  \tan(x) }

 =  >   \binom{lim}{x →0  }\frac{x( \cos(x)  +  \frac{ \sin(x) }{x}) }{x(x +  \frac{ \tan(x) }{x}) }

  =  > \frac{ \binom{lim}{x →0  } \cos(x) + \:  \binom{lim}{x →0  } \frac{ \sin(x) }{x}  }{  \binom{lim}{x →0  }x +  \binom{lim}{x →0  }  \frac{ \tan( x ) }{x}   }

 =  >  \frac{1 + 1}{0 + 1}

 =  >  \frac{2}{1}

 =  > 2

Answered by Anonymous
2

binom{lim}{x →0 }

 \binom{lim}{x →0  } \frac{x \cos(x) +  \sin(x)  }{ {x}^{2}  +  \tan(x) }

 =  >   \binom{lim}{x →0  }\frac{x( \cos(x)  +  \frac{ \sin(x) }{x}) }{x(x +  \frac{ \tan(x) }{x}) }

  =  > \frac{ \binom{lim}{x →0  } \cos(x) + \:  \binom{lim}{x →0  } \frac{ \sin(x) }{x}  }{  \binom{lim}{x →0  }x +  \binom{lim}{x →0  }  \frac{ \tan( x ) }{x}   }

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