I was trying to find asymptotes for a function f ( x ) = x e 1 x − 2

Question

I was trying to find asymptotes for a function   f  (  x  )  =  x      e                  1                  x          −          2

luisjoseblash2 · Accepted Answer

When x is large, let      1          x      −      2        =  t    ⟹    x  =  2  +      1    t      ⟹    x        e                  1                  x          −          2                      =            (      2      t      +      1      )        t        e    t  Now, using Taylor around t=0            (      2      t      +      1      )        t        e    t    =      1    t    +  3  +            5      t        2    +            7              t        2              6    +  O      (          t      3        )  Back to x  x        e                  1                  x          −          2                      =            6              x        3            −      18              x        2            +      15      x      +      1              6      (      x      −      2              )        2            Long division  x        e                  1                  x          −          2                      =  x  +  1  +      5          2      x        +  O      (          1              x        2              )  which gives not only the slant asymptote but also shows how the function does approach it.

Layla Velazquez · Accepted Answer

We have:      lim          x      →      +      ∞        f  (  x  )  =      lim          x      →      +      ∞        x  ⋅      e                  1                  x          −          2                      =  +  ∞We notice that       e                  1                  x          −          2                      →  1, so it&#039;s true the following asymptotic relation:  f  (  x  )      ∼      xwhen   x  →  +  ∞. This is a necessary but not sufficient condition for the existence of the asympot.We are searching for line of the type   y  =  x  +  k  ,    k  ∈      R        lim          x      →      +      ∞        f  (  x  )  −  x  =      lim          x      →      +      ∞        x  ⋅      (          e                        1                      x            −            2                                −    1    )        ∼          lim          x      →      +      ∞        x  ⋅      1          x      −      2        =  1Here, we have used a very important asymptotic relation:      e          f      (      x      )        −  1      ∼      f  (  x  )        x  →      x    0  With a function   f  (  x  ) such that   f  (  x  )  →  0 when   x  →      x    0    ∈                    R            ¯      Here, I put   f  (  x  )  =      1          x      −      2       that is such that   f  (  x  )  →  0 when   x  →  +  ∞In conclusion, the asympot is:  y  =  x  +  1

I was trying to find asymptotes for a function f ( x ) = x e <mrow class="MJ

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New Questions in Calculus 1