Question #1f1f7

1 Answer
Sep 30, 2017

#lim_(xrarroo)sqrt(x+sqrt(x+sqrtx))/sqrt(x+1)=1#

Explanation:

Let's try to factor within the square roots:

#lim_(xrarroo)sqrt(x+sqrt(x+sqrtx))/sqrt(x+1)#

#=lim_(xrarroo)sqrt(x+sqrt(x(1+1/sqrtx)))/sqrt(x(1+1/x))#

Pull out what we've just factored from their respective square roots:

#=lim_(xrarroo)sqrt(x+sqrtxsqrt(1+1/sqrtx))/(sqrtxsqrt(1+1/x))#

Factor from the numerator again:

#=lim_(xrarroo)sqrt(x(1+1/sqrtxsqrt(1+1/sqrtx)))/(sqrtxsqrt(1+1/x))#

And pull this from the numerator:

#=lim_(xrarroo)(sqrtxsqrt(1+1/sqrtxsqrt(1+1/sqrtx)))/(sqrtxsqrt(1+1/x))#

Which cancels:

#=lim_(xrarroo)sqrt(1+1/sqrtxsqrt(1+1/sqrtx))/sqrt(1+1/x)#

Note that both #1/x# and #1/sqrtx# approach #0# as #x# approaches infinity.

#=sqrt(1+0sqrt(1+0))/sqrt(1+0)#

#=sqrt1/sqrt1#

#=1#