What is the limit of # (x +sqrt((x^2)+(3x)))# as x goes to infinity?

Question

Luke Alvarez · Accepted Answer

#lim_(xrarroo) (x +sqrt(x^2+3x)) = oo#

#(x +sqrt(x^2+3x)) = x +sqrt(x^2(1+3/x))# for #x != 0#

# = x +sqrt(x^2)sqrt(1+3/x))# for #x != 0#

# = x +absx sqrt(1+3/x))# for #x != 0#

Now, as #xrarroo#, we have

#absx rarroo# and

#sqrt(1+3/x) rarr sqrt(1+0) = 1#

So,

#lim_(xrarroo) (x +sqrt(x^2+3x)) = lim_(xrarroo) (x +absx sqrt(1+3/x)) = oo#

Bonus

As #x# decreases without bound, the limit is quite different.

As #x rarr -oo#, we get indeterminate form: #-oo+oo#.

We think of #(x +sqrt(x^2+3x)) # as a ratio (over #1#) and eliminate the square root from the numerator.

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

# = (-3x)/(x-absx sqrt(1+3/x))#.

For #x < 0#, we have #absx = -x#, so we have:

#lim_(xrarr-oo)(x +sqrt(x^2+3x)) = lim_(xrarr-oo)(-3x)/(x-absx sqrt(1+3/x))#

# = lim_(xrarr-oo)(-3x)/(x+xsqrt(1+3/x))#

# = lim_(xrarr-oo)(-3)/(1+sqrt(1+3/x))#

# = -3/2#.

Aurora Alvarado · Answer

#lim_(x->+oo) (x+sqrt(x^2+3x)) = +oo#
#lim_(x->-oo) (x+sqrt(x^2+3x)) = -3/2# #sqrt(x^2+3x) >= 0# for all #x >= 0# So #lim_(x->+oo) (x+sqrt(x^2+3x)) >= lim_(x->+oo) x = +oo# Much more interesting is the case #lim_(x->-oo) (x+sqrt(x^2+3x))# First notice that #lim_(t->oo) (t - sqrt(t^2-C)) = 0# for any constant #C > 0# To see this: Given #epsilon > 0# #(t-epsilon)^2 = t^2 - 2tepsilon + epsilon^2# So if #t > (C+epsilon^2)/(2epsilon)#, then #(t-epsilon)^2 = t^2 - 2tepsilon + epsilon^2 < t^2 - (C +epsilon^2) + epsilon^2 = t^2 - C# So #t-epsilon < sqrt(t^2-C) < sqrt(t^2) = t# Let #t = -(x+3/2)# and #C = 9/4# Then: #0 = lim_(t->oo) (sqrt(t^2-C) - t)# #= lim_(x->-oo) (sqrt((x+3/2)^2 - 9/4) + x + 3/2)# #= lim_(x->-oo) (sqrt(x^2+3x+9/4 - 9/4) + x + 3/2)# #= lim_(x->-oo) (sqrt(x^2+3x) + x + 3/2)# #= lim_(x->-oo) (sqrt(x^2+3x) + x) + 3/2# Hence: #lim_(x->-oo) (sqrt(x^2+3x) + x) = -3/2#

Paisley Johnson · Answer

undefined The limit of (x + sqrt(x^2 + 3x)) as x goes to infinity is infinity.