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

Question

Isaiah Allen · Accepted Answer

Change the form of the expression.

#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#

Charles Arocha · Answer

undefined To find the limit of (sqrt(x^2 – 20x+3) – x) as x approaches infinity, we can simplify the expression by multiplying the numerator and denominator by the conjugate of the expression inside the square root, which is (sqrt(x^2 – 20x+3) + x). This will help eliminate the square root.

After simplifying, we get the expression (x^2 – 20x+3 - x^2) / (sqrt(x^2 – 20x+3) + x).

Simplifying further, we have (-20x+3) / (sqrt(x^2 – 20x+3) + x).

As x approaches infinity, the term -20x becomes insignificant compared to x, and the expression simplifies to 3 / (sqrt(x^2) + x).

Since the square root of x^2 is equal to x for positive values of x, the expression further simplifies to 3 / (x + x).

Finally, the limit of (sqrt(x^2 – 20x+3) – x) as x approaches infinity is 3 / (2x).