Someone please verify my answer!! ( limits)

Question

lim as x tends to positive infinity sqrt(3x^5-2x)-sqrt(3x^5-7x) 

 

Answer I got : (5)/(2√3)

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):

 

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

 

Please double check that you have typed the original expression correctly because it looks like you were on the right track and your answer would result for a power other than 5 inside the sqrt, that is if it were x^2. So you were definitely on the right track or maybe even just typed it wrong here?