How do you find the limit of #(lnx)^2/sqrtx# as #x-oo#?

Question

How do you find the limit of #(lnx)^2/sqrtx# as #x->oo#?

Emma Aldridge · Accepted Answer

#lim_(x->oo) (lnx)^2/sqrt(x) = 0#

The limit:

#lim_(x->oo) (lnx)^2/sqrt(x)#

is in the indeterminate form: #oo/oo# so we can use l'Hospital's rule:

#lim_(x->oo) (lnx)^2/sqrt(x) = lim_(x->oo) (d/dx(lnx)^2)/(d/dxsqrt(x)) = lim_(x->oo) (2lnx)/x 1/(1/2x^(3/2)) = lim_(x->oo) (4lnx)/x^(5/2)#

and again:

#lim_(x->oo) (4lnx)/x^(5/2) = 4 lim_(x->oo) (d/dx lnx)/(d/dx x^(5/2)) = 4 lim_(x->oo) 1/x 1/(5/2x^(7/2)) = 8/5 lim_(x->oo) 1/(x^(9/2)) = 0#

In general we have that for every #alpha > 0#:

#lim_(x->oo) lnx/x^alpha = lim_(x->oo) (d/dxlnx)/(d/dx x^alpha) = lim_(x->oo) 1/x 1/(alphax^(alpha-1)) = 1/alpha lim_(x->oo) 1/x^alpha = 0#

so: #lnx = o(x^alpha)#

So for any #a,b > 0#:

#lim_(x->oo) (lnx)^a/x^b = lim_(x->oo) (lnx/x^(b/a))^a =(lim_(x->oo) lnx/x^(b/a))^a = 0#

Emma Aldridge · Answer

undefined To find the limit of (lnx)^2/sqrtx as x approaches infinity, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get:

d/dx (lnx)^2 = 2lnx/x
d/dx sqrtx = 1/(2sqrtx)

Now, we can rewrite the expression as:

lim(x->oo) (2lnx/x) / (1/(2sqrtx))

Simplifying further:

lim(x->oo) (4lnx) / x^(3/2)

As x approaches infinity, the numerator (4lnx) grows without bound, while the denominator (x^(3/2)) also grows without bound. Therefore, the limit of (lnx)^2/sqrtx as x approaches infinity is infinity.

How do you find the limit of #(lnx)^2/sqrtx# as #x-&gt;oo#?

How do you find the limit of #(lnx)^2/sqrtx# as #x->oo#?