How do I do this limit (without hopital)

Question

limx--> 0⁺ (1/sqrtx) - ( 1/sqrtx^2+x)

Please show me steps. Im really confused. 

Answer 1

Use a common denominator, subtract, examine numerator and denominator.

Answer 2

We first combine fractions, rationalize the numerator, and then compare to the function with the x^2 dropped out instead of dropping out the x's in the denominator. This is because x goes to zero slower than x^2 in the case of a limit at 0, so it gives us a better chance at getting to 0:

x²/[ √x√(x²+x) (√(x²+x) + √x) ] < (√x)/2 by the fact that √(x²+x) > √x

and the expression on the right goes to zero as x goes to 0+. By the squeeze theorem, so does the original expression, using the fact that it is squeezed on the left side by the constant 0 function (everything is positive; we can see that in the original expression, or after we rationalize).

 

Or:

 

Switching it to a limit at (positive) infinity, the following limit is equivalent

 

lim_x→∞√x - x/√(x+1)      replaced each x with 1/x, then combined the fraction in the radical in the right denominator

 

(√x√(x+1) - x)/√(x+1)    combined fractions

 

= x/[√(x+1) * (√x√(x+1) + x)]   rationalized the numerator by multiplying top and bottom by the conjugate of the numerator

 

< x/[ √x(√x√x + x) ]     because √x < √(x+1), and a smaller denominator makes the overall fraction larger, for positive fractions

 

= 1/[ 2√x ]      by algebra

which goes to 0 as x goes to infinity.

 

By the squeeze theorem (using the fact that we can squeeze on the left by the constant 0 function: the function is always positive as x goes to infinity) the limit is zero. 

 

That connects it to an earlier problem.