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

Question

Isabella Ackerman · Accepted Answer

Use #(sqrta-b)(sqrta+b) = a-b^2# and #sqrt(u^2) = absu# and some other algebra.

#sqrt(x^2+x) -x = ((sqrt(x^2+x) -x))/1*((sqrt(x^2+x) +x))/((sqrt(x^2+x) +x)) # # = (x^2+x-x^2)/(sqrt(x^2+x) +x)# # = x/(sqrt(x^2(1+1/x)) +x)# #" "# (for #x!= 0#) # = x/(sqrt(x^2)sqrt(1+1/x) +x)# When evaluating the limit as #x# increases without bound, we are concerned only with positive values of #x#. For positive #x#, we have #sqrt(x^2) = x#, so, #sqrt(x^2+x) -x = x/(xsqrt(1+1/x) +x)# # = x/(x(sqrt(1+1/x) +1)# # = 1/(sqrt(1+1/x) +1)# So, finally, we get: #lim_(xrarroo)(sqrt(x^2+x) -x) = lim_(xrarroo) 1/(sqrt(1+1/x) +1)# # = 1/(sqrt(1+0) +1) = 1/2#

Paisley Johnson · Answer

undefined To find the limit of [(x^2+x)^(1/2)-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 [(x^2+x)^(1/2)+x]. This will help eliminate the square root in the numerator. Simplifying further, we get [(x^2+x) - x^2] / [(x^2+x)^(1/2)+x]. Canceling out the x^2 terms, we are left with x / [(x^2+x)^(1/2)+x]. As x approaches infinity, the x term dominates the expression, and the denominator becomes insignificant compared to x. Therefore, the limit of [(x^2+x)^(1/2)-x] as x approaches infinity is 1.