solve the following limit without L'Hospital rule or series: lim x → 0 arctan ⁡ 2 x sin ⁡ 3 x

Question

solve the following limit without L&#039;Hospital rule or series:      lim          x      →      0                  arctan      ⁡              2        x                    sin      ⁡              3        x

Johnathan Morse · Accepted Answer

First you want to show that      lim          y      →      0                  arctan      ⁡      y        y    =  1To do this, I would let   θ  =  arctan  ⁡  y. Then   θ  →  0 as   y  →  0, so      lim          y      →      0                  arctan      ⁡      y        y    =      lim          θ      →      0            θ          tan      ⁡      θ        =      lim          θ      →      0            θ          sin      ⁡      θ        ⋅  cos  ⁡  θ  =  1  ⋅  1  =  1(the limit       lim          θ      →      0                  sin      ⁡      θ        θ    =  1 is one of the basic limits you can usually invoke without proof).Now            arctan      ⁡      2      x              sin      ⁡      3      x        =                              arctan          ⁡          2          x                          2          x                                              sin          ⁡          3          x                          3          x                      ⋅      2    3

solve the following limit without L'Hospital rule or series: <munder> <mo movablelimits="t

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