How do you find the limit of #(sqrt(x^2 – 20x+3) – x)# as x approaches infinity?

1 Answer
Jim H
Feb 1, 2016

Change the form of the expression.

Explanation:

#sqrt(x^2 – 20x+3) – x = ((sqrt(x^2 – 20x+3) – x))/1*((sqrt(x^2 – 20x+3) + x)) /((sqrt(x^2 – 20x+3) +x)) #

# = ((x^2-20x+3)-x^2)/(sqrt(x^2 – 20x+3) +x)#

# = (-20x+3)/(sqrt(x^2 – 20x+3) +x)#

# = (-20x+3)/(sqrt(x^2)sqrt(1 – 20/x+3/x^2) +x)# #" "# (for #x != 0#)

# = (-20x+3)/(xsqrt(1 – 20/x+3/x^2) +x)# #" "# (for #x > 0#)

# = (cancel(x)(-20+3/x))/(cancel(x)(sqrt(1 – 20/x+3/x^2) +1)# #" "# (for #x > 0#)

As #x# increases without bound, the numerator approaches #-20# and the denominator approaches #sqrt1 + 1 = 2#.

Therefore,

#lim_(xrarroo) (sqrt(x^2 – 20x+3) – x ) = lim_(xrarroo)((x^2-20x+3)-x^2)/(sqrt(x^2 – 20x+3) +x)#

# = lim_(xrarroo) (cancel(x)(-20+3/x))/(cancel(x)(sqrt(1 – 20/x+3/x^2) +1)#

# = (-20)/(1+1) = -10#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
2326 views around the world
You can reuse this answer
Creative Commons License