Calculating slant asymptotes of radical functionI'm trying to calculate the slant asymptotes of the function x 2 + 2 x + 2 . I've found out that the gradients of the asymptotes are 1 for x → + ∞ and -1 for x → − ∞. I've also found out that the constant of the positive asymptote is 1. Intuitively, I know the constant of the negative asymptote is -1, but I'm struggling to show it through calculation. I need to evaluate this to find it: lim x → − ∞ x 2 + 2 x + 2 + xwithout using l'Hôpital's rule (for the purposes of the assignment I'm not supposed to know how to use it.) I have tried rationalizing the numerator using the conjugate but I just end up with an undefined value.

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Calculating slant asymptotes of radical functionI&#039;m trying to calculate the slant asymptotes of the function             x      2        +    2    x    +    2  . I&#039;ve found out that the gradients of the asymptotes are 1 for   x  →  +  ∞ and -1 for   x  →  −  ∞. I&#039;ve also found out that the constant of the positive asymptote is 1. Intuitively, I know the constant of the negative asymptote is -1, but I&#039;m struggling to show it through calculation. I need to evaluate this to find it:      lim          x      →      −      ∞                  x      2        +    2    x    +    2    +  xwithout using l&#039;Hôpital&#039;s rule (for the purposes of the assignment I&#039;m not supposed to know how to use it.) I have tried rationalizing the numerator using the conjugate but I just end up with an undefined value.

silenthunter440 · Accepted Answer

Hint:       x    2    +  2  x  +  2  =  (  x  +  1      )    2    +  1  ,, and       (    stuff          )      2        =      |    stuff      |    ,|, so for   x  ≠  −  1  ,, we have            x      2        +    2    x    +    2    =      (    x    +    1          )      2        +    1    =      |    x  +  1      |    ⋅      1    +          1              (        x        +        1                  )          2                      .If you want to evaluate that limit directly, it sounds like you got off to the right start by noting that             x      2        +    2    x    +    2   is never equal to x for real x, so we can certainly say that                                          x            2                    +          2          x          +          2                +        x                            =                  (                                    x              2                        +            2            x            +            2                    +          x          )                ⋅                                                            x                2                            +              2              x              +              2                        −            x                                                              x                2                            +              2              x              +              2                        −            x                                                            =                                            (                                                x                  2                                +                2                x                +                2                            +              x              )                                      (                                                x                  2                                +                2                x                +                2                            −              x              )                                                                          x                2                            +              2              x              +              2                        −            x                                                            =                                                            (                                                      x                    2                                    +                  2                  x                  +                  2                                )                            2                        −                          x              2                                                                          x                2                            +              2              x              +              2                        −            x                                                            =                              2            x            +            2                                                              x                2                            +              2              x              +              2                        −            x                                                            =                              2            x                                                              x                2                            +              2              x              +              2                        −            x                          +                  2                                                    x                2                            +              2              x              +              2                        −            x                          .            Now, noting that      lim          x      →      ∞                  x      2        +    2    x    +    2    −  x  =  +  ∞  ,it follows that      lim          x      →      −      ∞                  x      2        +    2    x    +    2    +  x  =      lim          x      →      −      ∞                  2      x                                x          2                +        2        x        +        2            −      x        =  2      lim          x      →      −      ∞                                                        1                                                                              1            x                                              x              2                        +            2            x            +            2                    −          1                      .But for x&amp;lt;0, we have                              1          x                                      x            2                    +          2          x          +          2                                    =        −                  1                                    |                        x                          |                                                            x            2                    +          2          x          +          2                                                  =        −                  1                                    x              2                                                            x            2                    +          2          x          +          2                                                  =        −                  1          +                      2            x                    +                      2                          x              2                                      .            .Can you take it from there?

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