Question 1044892
Your expression/function of {{{x}}} is
{{{f(x)=(x^"5/2"-2^"5/4")/(sqrt (x)-2^"1/4")}}}{{{"="}}}{{{(x^"5/2"-2^"5/4")/(x^"1/2"-2^"1/4")}}}{{{"="}}}{{{((x^"1/2")^5-(2^"1/4")^5)/(x^"1/2"-2^"1/4")}}} .
Looks complicated, but it can be simplified.
We need a change of variable, so we do not have to struggle with those fractional exponents.
 
If I rewrite your function with {{{x^"1/2"=a}}} ,
and define a constant {{{b=2^"1/4"}}} ,
{{{f(x)}}} looks as simple as {{{g(a)=(a^5-b^5)/(a-b)}}} , a rational function of {{{a}}} ,
with the constant {{{b}} such that {{b=2^"1/4"}}}-->{{{system(b^2=2^"1/2"=sqrt(2),"and",b^4=2)}}} .
Now, if {{{a<>b}}} <---> {{{a-b<>0}}} ,
we know a way to transform that rational function into a polynomial:
{{{g(a)=(a^5-b^5)/(a-b)=a^4+a^3b+a^2b^2+ab^3+b^4}}} for {{{a<>b}}} .
 
 
Now {{{f(x)=g(a)}}} , and {{{lim(x->sqrt(2),(x^"5/2"-2^"5/4")/(sqrt (x)-2^"1/4"))}}}{{{"="}}}
{{{lim(x->sqrt(2),f(x))}}}{{{"="}}}{{{lim(a->b,g(a))}}}{{{"="}}}{{{g(b)}}}{{{"="}}}{{{(b^4+b^3b+b^2b^2+b)b^3+b^4}}}{{{"="}}}{{{b^4+b^4+b^4b4+b^4+b^4}}}{{{"="}}}{{{5b^4=5*2=highlight(10)}}}


NOTES:
1) Your instructor may favor {{{u}}} rather than {{{a}}} as the variable for a change of variable.
I chose {{{a}}} because it should remind you of what you learned in algebra and/or "pre-calculus".
2) You may remember from algebra/pre-calculus that
{{{a^2-b^2=(a+b)(a-b)}}} , and
{{{a^3-b^3=(a^2+ab+b^2)(a-b)}}} .
You should not be surprised to find out that
{{{a^5-b^5=(a^4+a^3b+a^2b^2+ab^3+b^4)(a-b)}}} <---> {{{(a^5-b^5)/(a-b)=a^4+a^3b+a^2b^2+ab^3+b^4}}} .
Alternately, you may recognize {{{a^4+a^3b+a^2b^2+ab^3+b^4}}}
as the sum of the first five terms of a geometric sequence
with first term {{{a^4}}} and common ratio {{{r=b/a}}} ,
and that sum would be {{{a^4(r^5-1)/(r-1)=a^4(b^5/a^5-1)/(b/a-1)=a^4((b^5-a^5)/a^5)/((b-a)/a)=((b^5-a^5)/a)/((b-a)/a)=(b^5-a^5)/(b-a)=(a^5-b^5)/(a-b)}}}