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

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Answer 1 · 2016-04-15T15:36:00

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#