graphing f ( x ) = x ln ⁡ ( 1 + 1 x ) I was assigned to draw the graph of this function f ( x ) = x ln ⁡ ( 1 + 1 x ) When I calculate lim x → ∞ f ( x ) but the teacher said it's not correct even though its graph on the internet shows that lim x → ∞ f ( x ) = 1Please tell me where did I go wrong?

Question

graphing   f  (  x  )  =  x  ln  ⁡      (    1    +          1      x        )  I was assigned to draw the graph of this function   f  (  x  )  =  x  ln  ⁡      (    1    +          1      x        )  When I calculate       lim          x      →      ∞        f  (  x  ) but the teacher said it&#039;s not correct even though its graph on the internet shows that       lim          x      →      ∞        f  (  x  )  =  1Please tell me where did I go wrong?

last99erib · Accepted Answer

I am not sure of what you know but for small values of   x, we can use a Taylor expansion like so :  ln  ⁡  (  1  +  y  )  ≈  y  +            y      2        2    +  .  .  .Now let       1    x    =  y, when   x  →  ∞, you have   y  →  0 therefore you can write the following :  ln  ⁡  (  1  +      1    x    )  ≈      1    x    +      1          2              x        2              +  .  .  .So when you multiply by   x you get :  x  ln  ⁡  (  1  +      1    x    )  ≈  1  +      1          2      x        +  .  .  .Where   .  .  . are other power of       1    x   wich all tend to   0 as   x  →  ∞It should give you a good idea to what the limit tends to although a little bit more work is needed.

Layla Velazquez · Accepted Answer

First: You are right! The limit   1 is correct! To prove       lim          x      →      ∞        f  (  x  )  =  1 you could use L&#039;Hôpital&#039;s ruleIt holds:      lim          x      →      ∞        f  (  x  )  =      lim          x      →      ∞                  ln      ⁡      (      1      +      1              /            x      )              1              /            x        =      lim          x      →      ∞                  −              1                              x            2                    +          x                            −              1                  x          2                      =      lim          x      →      ∞                  x      2                      x        2            +      x        =      lim          x      →      ∞        1  −      1          x      +      1        =  1

graphing f ( x ) = x ln ⁡ ( 1 +

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