How do you find the limit #lnx/sqrtx# as #x->oo#?

Answer 1

#lim_(x->oo) lnx/sqrt(x) = 0#

As we have:

#lim_(x->oo) lnx = +oo# #lim_(x->oo) sqrt(x) = +oo#

The limit:

#lim_(x->oo) lnx/sqrt(x) = (+oo)/(+oo)#
presents itself in the indeterminate form #oo/oo# and we can use l'Hospital's rule:
#lim f(x)/g(x) = lim (f'(x))/(g'(x))#

so:

#lim_(x->oo) lnx/sqrt(x) = lim_(x->oo) (d/(dx)lnx)/(d/(dx)sqrtx) = lim_(x->oo) (1/x)/(1/(2sqrt(x))) = lim_(x->oo) (2sqrt(x))/x=lim_(x->oo) 2/sqrt(x) = 0#
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Answer 2

To find the limit of ln(x)/sqrt(x) as x approaches infinity, we can use L'Hôpital's rule. Taking the derivative of the numerator and denominator separately, we get (1/x)/(1/2sqrt(x)). Simplifying this expression, we have 2sqrt(x)/x. As x approaches infinity, the term 2sqrt(x) grows faster than x, so the limit of ln(x)/sqrt(x) 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.

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.

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.

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.

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