Show that lim x → 0 x ( 1 + cos ⁡ ( x ) ) − 2 tan ⁡ ( x ) 2 x − sin ⁡ ( x ) − tan ⁡ ( x ) = 7?

Question

Show that       lim          x      →      0                  x      (      1      +      cos      ⁡      (      x      )      )      −      2      tan      ⁡      (      x      )              2      x      −      sin      ⁡      (      x      )      −      tan      ⁡      (      x      )        =  7?

Jase Powell · Accepted Answer

By L&#039;Hôpital&#039;s rule we obtain:      lim          x      →      0                          x                  (          1          +          cos          ⁡                      x                    )                −        2        tan        ⁡                  x                            2        x        −        sin        ⁡                  x                −        tan        ⁡                  x                      =      lim          x      →      0                  1      +      cos      ⁡              x            −      x      sin      ⁡              x            −              2                              cos            2                    ⁡          x                            2      −      cos      ⁡              x            −              1                              cos            2                    ⁡          x                      =      lim          x      →      0                          cos        2            ⁡      x      +              cos        3            ⁡      x      −      x              cos        2            ⁡      x      sin      ⁡              x            −      2              2              cos        2            ⁡      x      −              cos        3            ⁡      x      −      1        =  =      lim          x      →      0            (                            cos          2                ⁡        x        +        2        cos        ⁡                  x                +        2                    1        +        cos        ⁡                  x                −                  cos          2                ⁡        x              −                  x                  cos          2                ⁡        x        sin        ⁡                  x                            (        cos        ⁡                  x                −        1        )        (        1        +        cos        ⁡                  x                −                  cos          2                ⁡        x        )              )    =  =      lim          x      →      0            (    5    −                  x        sin        ⁡                  x                            cos        ⁡                  x                −        1              )    =      lim          x      →      0            (    5    +                  x        sin        ⁡                  x                            2                  sin          2                ⁡                  x          2                      )    =      lim          x      →      0            (    5    +    2    ⋅                            x          2                cos        ⁡                              x            2                                      sin        ⁡                  x          2                      )    =  7.

dannyboi2006tk · Accepted Answer

The denominator seems simpler; the degree

1

expansions of sine and tangent are

\sin x = x + o (x), \tan x = x + o (x)

so

2 x - \sin x - \tan x = o (x)

, which doesn't help. So let's consider the expansion to degree

3

(they are odd functions, so the degree

2

term is zero):

\sin x = x - \frac{x^{3}}{6} + o (x^{3}), \tan x = x + \frac{x^{3}}{3} + o (x^{3})

so the denominator is

2 x - \sin x - \tan x = - \frac{x^{3}}{6} + o (x^{3})

Now expand the numerator

x (1 + \cos x) - 2 \tan x = x (1 + 1 - \frac{x^{2}}{2} + o (x^{2})) - 2 x - \frac{2 x^{3}}{3} + o (x^{3}) = - \frac{7}{6} x^{3} + o (x^{3})

Show that \lim_(x->0)(x(1+cos (x))-2\tan (x))/(2x-sin(x)-tan (x))=7?

Answered question

Answer & Explanation

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