#lim_(t->-oo) sqrt(t^2+2t+7 ) -sqrt(t^2-2t+5)#=?

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Answer 1 · 2018-04-13T16:59:10

The limit is #-2#

#lim_(t->-oo) sqrt(t^2+2t+7 ) -sqrt(t^2-2t+5)#

The initial form of the limit is indeterminate #oo-oo#.

Multiply by #(sqrt(t^2+2t+7 ) + sqrt(t^2-2t+5))/(sqrt(t^2+2t+7 ) + sqrt(t^2-2t+5)#, to get

#sqrt(t^2+2t+7 ) -sqrt(t^2-2t+5) = ((t^2+2t+7 ) - (t^2-2t+5))/( sqrt(t^2+2t+7 ) + sqrt(t^2-2t+5))#

# = (4t + 2)/( sqrt(t^2+2t+7 ) -sqrt(t^2-2t+5))#

# = (4t + 2)/( sqrt(t^2)sqrt(1+2/t+7/t^2 ) + sqrt(t^2)sqrt(1-2/t+5/t^2))# #" "# (for #t != 0#)

Now recall/note that #sqrt(t^2) = abs(t) = {(t,"if",t >= 0),(-t,"if"t<0):}#, so,

for #t < 0#, we have

#sqrt(t^2+2t+7 ) - sqrt(t^2-2t+5) = (4t + 2)/( -tsqrt(1+2/t+7/t^2 ) -tsqrt(1-2/t+5/t^2)#

# = (t(4+2/t))/(-t(sqrt(1+2/t+7/t^2 ) + sqrt(1-2/t+5/t^2)))#

# = (-(4+2/t))/((sqrt(1+2/t+7/t^2 ) + sqrt(1-2/t+5/t^2)))#

And now,

#lim_(t->-oo) sqrt(t^2+2t+7 ) 1-sqrt(t^2-2t+5) = lim_(trarr-oo)-(4+2/t)/((sqrt(1+2/t+7/t^2 ) + sqrt(1-2/t+5/t^2)))#

# = -(4+ 0)/(sqrt(1+0+0)+sqrt(1+0+0))#

# = -4/2 = - 2.#

Here is the graph of the function:

graph{sqrt(x^2+2x+7 ) -sqrt(x^2-2x+5) [-23.46, 22.15, -11.5, 11.3]}