Math, asked by padmaruttala257, 11 months ago

solve to become the brainliest​

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Answered by ITzBrainlyGuy
3

Answer:

lim_{x  =  > 0}( \frac{ \sin(x)  - x +  \frac{ {x}^{3} }{6} }{ {x}^{5} } ) \\

since evaluating limit of the numerator and the denominator would result in an indeterminate form use L'Hopital's rule

then we get

lim_{x => 0}( \frac{ \frac{d}{dx} ( \sin(x) - x +  \frac{ {x}^{3} }{6}  }{ \frac{d}{dx} ( {x}^{5}) }  )

calculate the derivatives

 lim_{x =   > 0}( \frac{ \cos( x )  - 1 +  \frac{1}{2}  {x}^{2} }{ {5x}^{4} } )

write all the numerators above the common denominator

 lim_{x =  > 0}( \frac{2 \cos(x) - 2 +  {x}^{2}  }{ \frac{2}{5 {x}^{2} } } )

simplify the complex fraction

 lim_{x =  > 0}( \frac{2 \cos(x) - 2 +  {x}^{2}  }{10 {x}^{4} } )

use the L'Hopital's rule

 lim_{x =  > 0}( \frac{ \frac{d}{dx}(2 \cos(x)  - 2 {x}^{2} ) }{ \frac{d}{dx} (10 {x}^{2}) } )

calculate the derivatives

lim [-2sin(x)+2x /40x³]

x→0

reduce the fraction

lim [sin(x)-x/20x³]

x→0

use L'Hopital's rule

lim [d/dx(-sin(x)+x)/d/dx(20x³)]

x→0

lim [-cos(x)+1/60x²]

x→0

use the L'Hopital's rule

lim [d/dx(-cos(x)+1/d/dx(60x²)]

x→0

calculate the derivatives

lim (sin(x)/120x)

x→0

use L'Hopital's rule

lim [d/dx(sin(x))/d/dx(120x)]

x→0

now, we get

lim [cos(x)/120]

x→0

equate the limit

then , we get

=cos(0°)/120

=1/120

hope this helps you please mark it as brainliest

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