What am I doing wrong?( limits,infinity,without hopital)

Question

limx-->positive infty sqrt(3x^5-2x)-sqrt(3x^5-7x) 

 

I first rationalized the whole thing. After this , i divided the top and the bottom by x^5 but the answer I got is not the real answer. What is the simplest way pf solving this? What am I doing wrong ?

Answer 1

The expression tends to 0 as x --> pos infinity. To see that, think of the original expression as a fraction over 1. Then multiply top and bottom of the fraction by the same quantity: 

sqrt(3x^5-2x) + sqrt(3x^5-7x)

That will give a numerator with the sqrt's removed (because (a - b)(a + b) = a2 - b2):

3x^5 - 2x - (3x^5 - 7x)/[sqrt(3x^5-2x) + sqrt(3x^5-7x)]

Now in the numerator the 3x^5 and the -3x^5 negate each other leaving:

(-2x + 7x)/[sqrt(3x^5-2x) + sqrt(3x^5-7x)] =

5x/[sqrt(3x^5-2x) + sqrt(3x^5-7x)]

Now multiply top and bottom by 1/x^(5/2) (notice that you do not divide top and bottom by x^5 because those terms cancel; the highest term now is x^(5/2) and that's what we divide top and bottom by):

5(1/(x^(3/2))/[sqrt(3-2(1/(x^4)) + sqrt(3-7(1/(x^4))]

Now it should be clear that the numerator goes to 0 while the denominator goes to 2sqrt(3), so limit = 0