SOLUTION: Kindly help me solving it i tried my best but I couldn't lim {{{(x^(5/2)-2^(5/4))/(sqrt (x)-2^(1/4))}}} is equal to x➡{{{sqrt (2)}}} a)1/10 b)10 c)20 d)n

Algebra ->  Functions -> SOLUTION: Kindly help me solving it i tried my best but I couldn't lim {{{(x^(5/2)-2^(5/4))/(sqrt (x)-2^(1/4))}}} is equal to x➡{{{sqrt (2)}}} a)1/10 b)10 c)20 d)n      Log On


   

Question 1044892: Kindly help me solving it i tried my best but I couldn't
lim %28x%5E%285%2F2%29-2%5E%285%2F4%29%29%2F%28sqrt+%28x%29-2%5E%281%2F4%29%29 is equal to
x➡sqrt+%282%29
a)1/10
b)10
c)20
d)none of these
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Your expression/function of x is
f%28x%29=%28x%5E%225%2F2%22-2%5E%225%2F4%22%29%2F%28sqrt+%28x%29-2%5E%221%2F4%22%29%22=%22%28x%5E%225%2F2%22-2%5E%225%2F4%22%29%2F%28x%5E%221%2F2%22-2%5E%221%2F4%22%29%22=%22 .
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%5E%221%2F2%22=a ,
and define a constant b=2%5E%221%2F4%22 ,
f%28x%29 looks as simple as g%28a%29=%28a%5E5-b%5E5%29%2F%28a-b%29 , a rational function of a ,
with the constant b%7D%7D+such+that+%7B%7Bb=2%5E%221%2F4%22-->system%28b%5E2=2%5E%221%2F2%22=sqrt%282%29%2C%22and%22%2Cb%5E4=2%29 .
Now, if a%3C%3Eb <---> a-b%3C%3E0 ,
we know a way to transform that rational function into a polynomial:
g%28a%29=%28a%5E5-b%5E5%29%2F%28a-b%29=a%5E4%2Ba%5E3b%2Ba%5E2b%5E2%2Bab%5E3%2Bb%5E4 for a%3C%3Eb .


Now f%28x%29=g%28a%29 , and %22=%22
lim%28x-%3Esqrt%282%29%2Cf%28x%29%29%22=%22lim%28a-%3Eb%2Cg%28a%29%29%22=%22g%28b%29%22=%22%28b%5E4%2Bb%5E3b%2Bb%5E2b%5E2%2Bb%29b%5E3%2Bb%5E4%22=%22b%5E4%2Bb%5E4%2Bb%5E4b4%2Bb%5E4%2Bb%5E4%22=%225b%5E4=5%2A2=highlight%2810%29

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%5E2-b%5E2=%28a%2Bb%29%28a-b%29 , and
a%5E3-b%5E3=%28a%5E2%2Bab%2Bb%5E2%29%28a-b%29 .
You should not be surprised to find out that
a%5E5-b%5E5=%28a%5E4%2Ba%5E3b%2Ba%5E2b%5E2%2Bab%5E3%2Bb%5E4%29%28a-b%29 <---> %28a%5E5-b%5E5%29%2F%28a-b%29=a%5E4%2Ba%5E3b%2Ba%5E2b%5E2%2Bab%5E3%2Bb%5E4 .
Alternately, you may recognize a%5E4%2Ba%5E3b%2Ba%5E2b%5E2%2Bab%5E3%2Bb%5E4
as the sum of the first five terms of a geometric sequence
with first term a%5E4 and common ratio r=b%2Fa ,
and that sum would be