How do you find the limit of $x(a^(1/x)-1)$ as $x->oo$ ?

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Answer 1 · 2017-06-01T19:10:58

$log_e a$

$x(a^(1/x)-1)= (a^(0+1/x)-a^0)/(1/x)$ now calling $1/x = h$ we have

$lim_(x->oo)x(a^(1/x)-1)=lim_(h->0)(a^(0+h)-a^0)/h = a^0log_e a=log_e a$

Answer 2 · 2017-06-01T19:13:13

$ln(a)$

Rewrite the limit as

$lim_(x->oo)(a^(1/x)-1)/(1/x)$

So that it produces the indeterminate form $0/0$ .

Now use L'hopital's rule (and chain rule as a result):

$lim_(x->oo)(a^(1/x)-1)/(1/x)=lim_(x->oo)(a^(1/x) * ln(a) * (-1)/x^2)/((-1)/x^2)$

$=lim_(x->oo)a^(1/x)ln(a) = a^0ln(a) = ln(a)$

Final Answer

How do you find the limit of x(a^(1/x)-1)x(a^(1/x)-1) as x->oox->oo?