We have the following limits: lim x → π 4 1 − tan 2 ⁡ x cos ⁡ 2 x and lim θ → 1 = θ − 1 + sin ⁡ ( θ 2 − 1 ) θ 2 − 1 Using L'Hôpital I obtained that the second limit is θ = 3 2 . For the first limit I was able to rewrite it to lim x → π 4 ( 1 − tan 2 ⁡ ( x ) ) sec ⁡ ( 2 x ), but I don't really know what to do with that.

Question

We have the following limits:      lim          x      →              π        4                                1        −                  tan          2                ⁡        x                    cos        ⁡        2        x             and       lim          θ      →      1        =                    θ        −        1        +        sin        ⁡        (                  θ          2                −        1        )                              θ          2                −        1            Using L&#039;Hôpital I obtained that the second limit is   θ  =            3      2      . For the first limit I was able to rewrite it to       lim          x      →              π        4              (  1  −      tan    2    ⁡  (  x  )  )  sec  ⁡  (  2  x  ), but I don&#039;t really know what to do with that.

Marisol Morton · Accepted Answer

For the first limit, some elementary trigonometry (a duplication formula) will do:            1      −              tan        2            ⁡      x              c      o      s      2      x        =                    cos        2            ⁡      x      −              sin        2            ⁡      x                      cos        2            ⁡      x      cos      ⁡      2      x        =            cos      ⁡      2      x                      cos        2            ⁡      x      cos      ⁡      2      x        =      1                  cos        2            ⁡      x            →          x      →                        π          4                          2.For the second limit, use asymptotic analysis:            θ      −      1      +      sin      ⁡      (              θ        2            −      1      )                      θ        2            −      1        =            θ      −      1      +              θ        2            −      1      +      o      (              θ        2            −      1      )                      θ        2            −      1        =      1          θ      +      1        +  1  +  o  (  1  )      →          θ      →      1                1    2    +  1.

Jonathan Miles · Accepted Answer

use the       lim          x      →      0                  sin      ⁡      x        x    =  1                    θ        −        1        +        sin        ⁡        (                  θ          2                −        1        )                              θ          2                −        1              =            θ      −      1                      θ        2            −      1        +            sin      ⁡      (              θ        2            −      1      )                      θ        2            −      1        =      1          θ      +      1        +            sin      ⁡      (              θ        2            −      1      )                      θ        2            −      1            lim          x      →      1        (      1          θ      +      1        +            sin      ⁡      (              θ        2            −      1      )                      θ        2            −      1        )  =  1      /    2  +  1  =  3      /    2

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