What's wrong in my method to find values of a and b in the limit lim x → 0 x ( 1 + a cos ⁡ x ) − b sin ⁡ x sin 3 ⁡ x = 1

Question

What&#039;s wrong in my method to find values of  a  and  b  in the limit       lim          x      →      0                  x      (      1      +      a      cos      ⁡      x      )      −      b      sin      ⁡      x                      sin        3            ⁡      x        =  1

a2g1g9x · Accepted Answer

The correct method using L.H. rule will beFirst multiply and divide by       x    3   and separate       lim          x      →      0                  x      3                      sin        3            ⁡      x        =  1 to get      lim          x      →      0                  x      (      1      +      a      cos      ⁡      x      )      −      b      sin      ⁡      x              x      3        =  1    ⟹        lim          x      →      0                  1      −      (      b      −      a      )      cos      ⁡      x      −      a      x      sin      ⁡      x              3              x        2              =  1      lim          x      →      0                  1      −      (      b      −      a      )      cos      ⁡      x              x      2        −      lim          x      →      0                  a      sin      ⁡      x        x    =  3For the first limit to exist ,   b  −  a  =  1 which gives      1    2    −  a  =  3    ⟹    a  =  −      5    2      ⟹    b  =  −      3    2

What's wrong in my method to find values of a and b in the limit <munder> <mo movablelimi

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New Questions in Calculus 1