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

Question

Cameron Alexander · Accepted Answer

#= 0#

#lim_(x to oo) sqrt(x^2 + 1) - x#

#= lim_(x to oo) x( sqrt(1 + 1/x^2) - 1) #

#= lim_(x to oo) x( (1 + 1/x^2)^(1/2) - 1) #

Binomial expansion on the radical

#=lim_(x to oo) x( (1 + 1/2 *1/x^2 + O(1/x^4)) - 1) #

#=lim_(x to oo) x( 1/(2 x^2) + O(1/x^4)) #

#=lim_(x to oo) 1/(2 x) + O(1/x^3) #

#= 0#

Scarlett Allison · Answer

Rewrite the expression using the conjugate.

#((sqrt(x^2+1)-x))/1 * ((sqrt(x^2+1)+x))/((sqrt(x^2+1)+x)) = ((x^2+1)-x^2)/(sqrt(x^2+1)+x)# # = 1/(sqrt(x^2+1)+x)# As #x rarr oo#, the denominator also increases without bound, so the ratio goes to #0#. The following is optional. From # = 1/(sqrt(x^2+1)+x)#, we could continue: For #x > 0#, we get # = 1/(sqrt(x^2)sqrt(1+1/x^2)+x)# # = 1/(x sqrt(1+1/x^2)+x)# # = 1/(x(sqrt(1+1/x^2)+1)# As #x# increases without bound, #(sqrt(1+1/x^2)+1) rarr 2#, and #x rarr oo#, so #1/(x(sqrt(1+1/x^2)+1)) rarr 0#

Cameron Alexander · Answer

undefined To find the limit of sqrt(x^2 + 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 + 1) + x. This will help eliminate the square root in the numerator. After simplifying, we get (sqrt(x^2 + 1) - x) * (sqrt(x^2 + 1) + x) / (sqrt(x^2 + 1) + x). Simplifying further, we have (x^2 + 1 - x^2) / (sqrt(x^2 + 1) + x). This simplifies to 1 / (sqrt(x^2 + 1) + x). As x approaches infinity, the denominator also approaches infinity, so the limit of the expression is 1 / infinity, which is equal to 0. Therefore, the limit of sqrt(x^2 + 1) - x as x approaches infinity is 0.