# Evaluate. Justify. If limit does'nt exist explain. limx->∞ 7x^3+6x^2+1/(x+1)(2x^2+2) ; limx->-2 (x-1)sin(x+2)/x^2-4 ; limx->-∞ sqrtx+x^2/x

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`lim_(x->oo) (7x^3+6x^2+1)/((x+1)(2x^2+2))`

`=lim_(x->oo) (7x^3+6x^2+1)/(2x^3+2x^2+2x+2)`

Divide both top and bottom polynomials by x^3 and we get

`lim_(x->oo) (7+6/x+1/x^3)/(2 + 2/x + 2/x^2 + 2/x^3)`

Now we know `lim_(x->oo) 1/x^n = 0` for all n>0

So our limit is equal to `(7+0+0)/(2+0+0+0) = 7/2`

`lim_(x->oo) (7x^3+6x^2+1)/((x+1)(2x^2+2)) = 7/2`

`lim_(x->-2) ((x-1)sin(x+2))/(x^2-4)` `= lim_(x->-2) ((x-1)sin(x+2))/((x+2)(x-2))`

`=lim_(x->-2) (x+1)/(x-2) * lim_(x->-2) (sin(x+2))/(x+2)`

Substituting t=x+2 we get `lim_(t->0) (sin(t))/t` and we know this = 1, so

`lim_(x->-2) ((x-1)sin(x+2))/(x^2-4) = (-2+1)/(-2-2) * 1 = -1/-4 = 1/4` .

`lim_(x->-2) ((x-1)sin(x+2))/(x^2-4)` `= lim_(x->-2) ((x-1)sin(x+2))/((x+2)(x-2))`

`=lim_(x->-2) (x+1)/(x-2) * lim_(x->-2) (sin(x+2))/(x+2)`

Substituting t=x+2 we get `lim_(t->0) (sin(t))/t` and we know this = 1, so

`lim_(x->-2) ((x-1)sin(x+2))/(x^2-4) = (-2+1)/(-2-2) * 1 = -1/-4 = 1/4` .