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

Answer 1

Evelyn Agard

# 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#

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Answer 2

Isaiah Allen

#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

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Answer 3

Ezekiel Alexander

To find the limit of the expression x(a^(1/x) - 1) as x approaches infinity, we can use the concept of limits.

Let's simplify the expression first. We can rewrite it as x(a^(1/x) - 1) = x(a^(1/x) - 1) * (a^(1/x) + 1)/(a^(1/x) + 1).

Now, we can apply the limit properties. As x approaches infinity, the term (1/x) approaches 0. Therefore, we can rewrite the expression as x((a^(1/x) - 1) * (a^(1/x) + 1)/(a^(1/x) + 1)) = x((a^0 - 1) * (a^0 + 1)/(a^0 + 1)).

Simplifying further, we have x(0 * (a^0 + 1)/(a^0 + 1)) = x(0) = 0.

Hence, the limit of x(a^(1/x) - 1) as x approaches infinity is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #x(a^(1/x)-1)# as #x-&gt;oo#?

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