What is #lim_(x->oo) x - sqrt(x^2 + 4x + 3)#?

#y = lim_(x->oo) x - sqrt(x^2 + 4x + 3)#

The function #y = -sqrt(x^2 + 4x + 3)# opposes the function #y = x#, and approaches a value for #y# faster than #y = x# ever reaches a value, so we can expect the function to approach its limit from above; however, there is no clear indication of what value it happens to be. You cannot do #oo - oo# and get a finite answer or #oo# back. So, we need to look at the actual graph:

graph{x - sqrt(x^2 + 4x + 3) [-12.73, 19.3, -11.61, 4.41]}

Looking here, it's actually pretty clearly approaching #y = -2#. So:

#y = lim_(x->oo) x - sqrt(x^2 + 4x + 3) = -2#

This kind of problem is usually introduced after students have worked with limits at infinity of ratios involving similar expressions. The trick here is to turn this expression into a ratio whose limit we can evaluate.

#x-sqrt(x^2+4x+3) = (x-sqrt(x^2+4x+3))/1#

We'll use the (by now familiar) technique of rationalizing the numerator.

#((x-sqrt(x^2+4x+3)))/1 * ((x+sqrt(x^2+4x+3)))/((x+sqrt(x^2+4x+3))) = (x^2-(x^2+4x+3))/(x+sqrt(x^2+4x+3))#

#= (-4x-3)/(x+sqrt(x^2+4x+3))#

Now we'll use the fact that. for all #x != 0#,

#sqrt(x^2+4x+3) = sqrt(x^2(1+4/x+3/x^2)) = sqrt(x^2) sqrt(1+4/x+3/x^2)#

And we also need: #sqrt(x^2)= absx#, so for positive #x#, we have #sqrt(x^2) = x#

So for positive values of #x#, we get:

#x-sqrt(x^2+4x+3) = (-4x-3)/(x+sqrt(x^2+4x+3))#

# = (-4x-3)/(x+xsqrt(1+4/x+3/x^2))#

# = (x(-4-3/x))/(x(1+sqrt(1+4/x+3/x^2)))#

# = (-4-3/x)/(1+sqrt(1+4/x+3/x^2))#

#lim_(xrarroo)(-4-3/x)/(1+sqrt(1+4/x+3/x^2)) = (-4-0)/(1+sqrt(1+0+0))=(-4)/2=-2#