Evaluate lim x → 0 sin ⁡ ( x + tan ⁡ x ) − 2 x cos ⁡ x x ( sin ⁡ x 2 ) 2

Question

Evaluate      lim          x      →      0                  sin      ⁡      (      x      +      tan      ⁡      x      )      −      2      x      cos      ⁡      x              x      (      sin      ⁡              x        2                    )        2

Makai Blackwell · Accepted Answer

You have messed up with asymptotic relations. You must be really really careful when you use them.In this limit, we know from the denominator that we have to search for a 5 degree expansion. Namely:  x  ⋅  (  sin  ⁡  (      x    2    )      )    2        ∼          x    5  Now, we set things ready for Tyalor-MacLaurin series:  sin  ⁡  (  t  )  =  t  −      1    6        t    3    +      1    120        t    5    +  o  (      t    5    )  tan  ⁡  (  t  )  =  t  +      1    3        t    3    +      2    15        t    5    +  o  (      t    5    )  cos  ⁡  (  t  )  =  1  −      1    2        t    2    +      1    24        t    4    +  o  (      t    4    )Now, we can work on t numerator:  sin  ⁡  (  x  +  tan  ⁡  (  x  )  )  −  2  x  cos  ⁡  (  x  )  =  sin  ⁡      (    2    x    +          1      3              x      3        +          2      15              x      5        +    o    (          x      5        )    )    −  2  x      (    1    −          1      2              x      2        +          1      24              x      4        +    o    (          x      4        )    )    =  2  x  +      1    3        x    3    +      1    120        x    5    −      1    6              (      2      x      +              1        3                    x        3            +              2        15                    x        5            +      o      (              x        5            )      )        3    +      1    120              (      2      x      +              1        3                    x        3            +              2        15                    x        5            +      o      (              x        5            )      )        5    −  2  x  +      x    3    −      1    12        x    5    +  o  (      x    5    )  =  −      7    20        x    5    +  o  (      x    5    )Note that, here, I haven&#039;t done all the calculations in order to exapand the 3 and 5 powers of quadrinomials, but I have kept only the most significant (grade 1, 3 and 5).So:      lim          x      →      0                  sin      ⁡      (      x      +      tan      ⁡      x      )      −      2      x      cos      ⁡      x              x      (      sin      ⁡              x        2                    )        2              =      lim          x      →      0                  −              7        20                    x        5            +      o      (              x        5            )              x      5        =  −      7    20

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