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

2 Answers
Aug 24, 2016

Answer:

#= 0#

Explanation:

#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#

Aug 24, 2016

Answer:

Rewrite the expression using the conjugate.

Explanation:

#((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#