What is #lim_(x->oo) (sqrt(xsqrt(x+sqrt(x)))-sqrt(x))# ?

#lim_(x->oo) sqrt(xsqrt(x+sqrt(x))) - sqrtx#

If we examine the innermost #x + sqrtx#, we can plainly see that that will approach #oo# because #sqrt(oo + sqrtoo) -> sqrtoo#. As you increase a number, it approaches #oo#; that means that at large values, #sqrtoo = oo#.

Thus, for the left outermost term, we have:

#sqrt(xsqrt(x+sqrt(x))) => sqrt(x*oo)#

#= sqrt(oo)*sqrt(oo) -> oo#

Then, intuitively, the #-sqrt(oo)# approaches #-oo# for large #x#. So, we have some form of #oo - oo'#.

Finally, we realize that #sqrt(xsqrt(x+sqrt(x)))# grows towards #oo# faster than #sqrtx# grows towards #oo'#. You can see that because #sqrt(x+sqrtx)# >> #1# at large #x#.

Therefore, the subtraction gives more significance to the left #oo# term as #x# grows very large, and so the limit becomes:

#oo - oo' = oo - "smaller number" => color(blue)(oo)#.

When #x > 16# we have

#x+sqrt(x) > 16 + sqrt(16) = 16 + 4 = 20#

So

#sqrt(x+sqrt(x)) > sqrt(20) > 4#

So

#sqrt(x sqrt(x+sqrt(x))) > sqrt(4x) = 2sqrt(x)#

So

#sqrt(x sqrt(x+sqrt(x))) - sqrt(x) > 2sqrt(x) - sqrt(x) = sqrt(x)#

As #x->oo#, we know #sqrt(x) -> oo#

So

#sqrt(x sqrt(x+sqrt(x))) - sqrt(x) -> oo#

#sqrt(x sqrt(x+sqrt(x))) - sqrt(x) = sqrt(x (sqrt(x+sqrt(x)))) - sqrt(x) #

# = sqrtx sqrt((sqrt(x+sqrt(x)))) - sqrtx * 1#

# = sqrtx (sqrtsqrt(x+sqrt(x)) - 1)#

# = sqrtx (root(4)(x+sqrt(x)) - 1)#

As #x# increases without bound, both factors increase without bound, so the product increases without bound.

#lim_(xrarroo)(sqrt(x sqrt(x+sqrt(x))) - sqrt(x))= lim_(xrarroo)sqrtx(root(4)(x+sqrt(x)) - 1)#

#= (lim_(xrarroo)sqrtx)(lim_(xrarroo)(root(4)(x+sqrt(x)) - 1)#

# = oo*oo=oo#