I have the following: lim x → 0 1 − sin ⁡ ( x ) x log ⁡ ( 1 + 2 x 2 ) .I set 2 x 2 = t knowing that lim t → 0 log ⁡ ( 1 + t ) t = 1but I am not sure if I am doing it right.

Question

I have the following:      lim          x      →      0                  1      −                        sin          ⁡          (          x          )                x                    log      ⁡      (      1      +      2              x        2            )        .I set   2      x    2    =  t knowing that       lim          t      →      0                  log      ⁡      (      1      +      t      )        t    =  1but I am not sure if I am doing it right.

Amy Daniels · Accepted Answer

lim              x        →        0                            x        −        sin        ⁡        x                    x        ⋅        ln        ⁡        (        1        +        2                  x          2                )              =          lim              x        →        0                            x        −        sin        ⁡        x                    2                  x          3                      ×          lim              x        →        0                            2                  x          2                            ln        ⁡        (        1        +        2                  x          2                )              =          lim              x        →        0                            x        −        sin        ⁡        x                    2                  x          3                      ×    1  Above we have used      ∙          lim              y        →        0                            ln        ⁡        (        1        +        y        )            y        =    1  Now our limit convert into            lim              x        →        0                            x        −        sin        ⁡        x                    2                  x          3                    Now Using DL,HopitalRule3timesSo Limit            lim              x        →        0                            1        −        cos        ⁡        x                    6                  x          2                    So Limit            lim              x        →        0                            sin        ⁡        x                    12        x            So Limit            lim              x        →        0                            cos        ⁡        x            12        =          1      12

I have the following: <munder> <mo movablelimits="true" form="prefix">lim <mrow clas

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