Evalutating lim x → ∞ ( x + 1 ) k − ( x ) k , given 0 &lt; k &lt; 1

Question

Evalutating       lim          x      →      ∞        (  x  +  1      )    k    −  (  x      )    k  , given   0  &amp;lt;  k  &amp;lt;  1

haingear8v · Accepted Answer

You can just rewrite the expression and use L'Hôpital's rule:

lim_{x \to + \infty} ((x + 1)^{k} - x^{k}) = lim_{x \to + \infty} \frac{(1 + 1 / x)^{k} - 1}{1 / x^{k}} = lim_{x \to + \infty} \frac{(- k / x^{2}) (1 + 1 / x)^{k - 1}}{- k x^{- k - 1}} = lim_{x \to + \infty} (x + 1)^{k - 1} = 0

civilnogwu · Accepted Answer

Use Taylor polynomial: as   h  →  0, we have   (  1  +  h      )    k    =  1  +  k  h  +  o  (  h  )As   x  →  ∞  (  x  +  1      )    k    =      x    k              (      1      +              1        x            )        k    =      x    k        (    1    +          k      x        +    o    (    1          /        x    )    )    =      x    k    +  k      x          k      −      1        +  o  (      x          k      −      1        )    (  x  +  1      )    k    −      x    k    =  k      x          k      −      1        +  o  (      x          k      −      1        )Now   k  &amp;lt;  1 so   k  −  1  &amp;lt;  0 and thus       lim          x      →      ∞            x          k      −      1        =  0. Conclude  (  x  +  1      )    k    −      x    k    →  0     as   x  →  ∞or more pricisely  (  x  +  1      )    k    −      x    k    ∼  k      x          k      −      1           as   x  →  ∞Example:      x    +    1    −      x    ∼      1          2              x                 as   x  →  ∞

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