How to solve this limit involving sine and log? lim x → 0 + x 2 s i n 1 x ln ⁡ ( 1 + 2 x )

Question

How to solve this limit involving sine and log?      lim          x      →              0        +                                x        2            s      i      n              1        x                    ln      ⁡      (      1      +      2      x      )

Carter Escobar · Accepted Answer

0  ≤      |                            x          2                sin        ⁡                  1          x                            ln        ⁡        (        1        +        2        x        )              |    ≤            x      2              ln      ⁡      (      1      +      2      x      )        ∼            2      x              2              1        +        2        x              =  x  (  1  +  2  x  )  →  0, where ∼ is the result of applyin L&#039;Hospital.

Waylon Ruiz · Accepted Answer

Note that, by applying the Squeeze theorem you get                                  −        1        ≤        sin        ⁡        x        ≤        1                                          ⇒                  lim                      x            →            ∞                                                −            1                    x                ≤                  lim                      x            →            ∞                                                sin            ⁡            x                    x                ≤                  lim                      x            →            ∞                                    1          x                                                  ⇒        0        ≤                  lim                      x            →            ∞                                                sin            ⁡            x                    x                ≤        0                                          ⇒                  lim                      x            →            ∞                                                sin            ⁡            x                    x                =        0                                          ⇒                  lim                      y            →            0            +                          y        sin        ⁡                  1          y                =        0            so                              lim                      x            →                          0              +                                                                          x              2                        sin            ⁡                          1              x                                            ln            ⁡            (            1            +            2            x            )                                              =                  lim                      x            →                          0              +                                      sin        ⁡                  (                      1            x                    )                                                    x              2                        ⋅            2            x                                ln            ⁡            (            1            +            2            x            )            ⋅            2            x                                                            =                  lim                      x            →                          0              +                                      sin        ⁡                  (                      1            x                    )                          x          2                                                  =        0

How to solve this limit involving sine and log? <munder> <mo movablelimits="true" form="pr

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