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

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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]}

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