If limit is zero: lim x → 0 ( sin ⁡ 2 x x 3 + a x 2 + b ) = 0then find a+b=?

Question

If limit is zero:      lim          x      →      0            (                  sin        ⁡        2        x                    x        3              +          a              x        2              +    b    )    =  0then find a+b=?

Dustin Durham · Accepted Answer

Solution Using L&#039;Hospital&#039;s Rule:      lim          x      →      0            (                  sin        ⁡        2        x                    x        3              +          a              x        2              +    b    )    =      lim          x      →      0            (                  s        i        n        (        2        x        )        +        a        x        +        b                  x          3                            x        3              )    =      lim          x      →      0            (                  2        c        o        s        (        2        x        )        +        a        +        3        b                  x          2                            3                  x          2                      )  But We&#039;re given that the limit exists and is equal to zero hence we must have a+2=0 i.e.a=−2.Now as a=−2 therefore the following limit is zero      lim          x      →      0            (                  sin        ⁡        2        x                    x        3              −          2              x        2              +    b    )  Applying the same technique as above we get   b  =  4      /    3

Jeffery Clements · Accepted Answer

A Taylor&#039;s expansion shows:            sin      ⁡      2      x              x      3        +      a          x      2        +  b  =            a      +      2              x      2        +      (    b    −          4      3        )    +            4              x        2              15    −            8              x        4              315    +  O      (          x      6        )  Therefore   a  =  −  2 and   b  =      4    3  Some explanations:define   f  (  x  )  =      x    2    (            sin      ⁡      2      x              x      3        +      a          x      2        +  b  )  =            sin      ⁡      2      x        x    +  a  +  b      x    2  . Now recall   sin  ⁡  x  =  x  −      1    6        x    3    +      1    120        x    5    +  O  (      x    6    ). Therefore  f  (  x  )  =  a  +  b      x    2    +  2  −      4    3        x    2    +      4    15        x    4    +  O  (      x    6    )hence            sin      ⁡      2      x              x      3        +      a          x      2        +  b  =            a      +      b              x        2            +      2      −              4        3                    x        2            +              4        15                    x        4            +      O      (              x        6            )              x      2        =            a      +      2              x      2        +  (  b  −      4    3    )  +      4    15        x    2    +  O  (      x    4    )Now if you want the limit at zero to be zero you must have a=−2 so that the first term is zero and then   b  =      3    4  . the rest of the terms in the expansion are zero at zero, since they are a multiple of x.

If limit is zero: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="M

Answered question

Answer & Explanation

New Questions in Differential Calculus