Find lim x → 0 ⌊ tan ⁡ 2 x sin ⁡ x ⌋ My try: tan ⁡ 2 x = sin ⁡ 2 x cos ⁡ 2 x sin ⁡ 2 x = 2 cos ⁡ x sin ⁡ xSo: tan ⁡ 2 x sin ⁡ x = 2 cos ⁡ x cos ⁡ 2 x So: lim x → 0 ⌊ tan ⁡ 2 x sin ⁡ x ⌋ = lim x → 0 ⌊ 2 cos ⁡ x cos ⁡ 2 x ⌋

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Find       lim          x      →      0            ⌊                  tan        ⁡        2        x                    sin        ⁡        x              ⌋  My try:  tan  ⁡  2  x  =                    sin        ⁡        2        x                    cos        ⁡        2        x              sin  ⁡  2  x  =  2  cos  ⁡  x  sin  ⁡  xSo:                    tan        ⁡        2        x                    sin        ⁡        x              =                    2        cos        ⁡        x                    cos        ⁡        2        x            So:      lim          x      →      0            ⌊                            tan          ⁡          2          x                          sin          ⁡          x                      ⌋    =      lim          x      →      0            ⌊                            2          cos          ⁡          x                          cos          ⁡          2          x                      ⌋

Maeve Holloway · Accepted Answer

x  ≠  0  and around 0 we have            2      cos      ⁡              x                    cos      ⁡      2      x        &amp;gt;  2because it&#039;s            2      cos      ⁡              x                    2              cos        2            ⁡      x      −      1        &amp;gt;  2or  (  1  −  cos  ⁡      x    )  (  1  +  2  cos  ⁡      x    )  &amp;gt;  0.Thus,      lim          x      →      0            [                  2        cos        ⁡                  x                            cos        ⁡        2        x              ]    =  2

Leon Robinson · Accepted Answer

Let&#039;s consider  f  (  x  )  =            2      cos      ⁡      x              cos      ⁡      2      x      over   (  −  π      /    2  ,  π      /    2  ). Then      f    ′    (  x  )  =  2            −      sin      ⁡      x      cos      ⁡      2      x      +      2      cos      ⁡      x      sin      ⁡      2      x                      cos        2            ⁡      2      x        =  2            sin      ⁡      x      (      2              cos        2            ⁡      x      +      1      )                      cos        2            ⁡      2      x      Therefore f has a local minimum at 0 and f(0)=2.In a suitable neighborhood of 0 we have   2  ≤  f  (  x  )  &amp;lt;  3

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New Questions in Calculus 1