How do you determine the limit of #(sqrt(x^2+10x+1)-x)# as x approaches infinity?

Question

Charles Arocha · Accepted Answer

5

#sqrt(x^2+10x+1)-x# #= x { (sqrt(1+10/x+1/x^2)-1}# because #lim_{x o oo} 10/x+1/x^2 = 0# .... we can use a Binimial Expansion on #sqrt(1+10/x+1/x^2)# as follows #sqrt(1+[10/x+1/x^2])# #= 1 + 1/2 [10/x+1/x^2] + (1/2 (- 1/2))/(2!) [10/x+1/x^2]^2# #approx 1 + 5/x + mathcal{O} (1/x^2)# so #lim_{x to oo} x { (sqrt(1+10/x+1/x^2)-1}# #= lim_{x to oo} x { 1+5/x-1}# #=5#

Luke Alvarez · Answer

undefined To determine the limit of (sqrt(x^2+10x+1)-x) as x approaches infinity, we can simplify the expression by multiplying both the numerator and denominator by the conjugate of the expression, which is (sqrt(x^2+10x+1)+x). This will help eliminate the square root.

After simplifying, we get the expression (x^2+10x+1 - x^2) / (sqrt(x^2+10x+1)+x).

Simplifying further, we have (10x+1) / (sqrt(x^2+10x+1)+x).

As x approaches infinity, the term 10x becomes dominant, and the expression can be simplified to 10x / (x + x).

This further simplifies to 10 / 2, which equals 5.

Therefore, the limit of (sqrt(x^2+10x+1)-x) as x approaches infinity is 5.