Question #72096

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Answer 1 · 2017-08-23T17:48:58

The limit equals $3$ .

Divide by the highest power, which is $x^2$ .

$L = lim_(x->oo) ((3x^2 + 3)/x^2)/((x^2 - 7x +9 )/x^2)$

$L = lim_(x->oo) (3 + 3/x^2)/(1 - 7/x + 9/x^2)$

We know that $lim_(x->oo) 1/x = 0$ , therefore:

$L = 3/1$

$L = 3$

Hopefully this helps!

Answer 2 · 2017-08-23T17:58:12

Because the expression evaluated at $oo$ yields the indeterminate form $oo/oo$ , one should use L'Hôpital's rule.

Given: $lim_(x to oo)(3x^2+3)/(x^2-7x+9)$

To apply L'Hôpital's rule, differentiate the numerator and the denominator:

$lim_(xtooo)((d(3x^2+3))/dx)/((d(x^2-7x+9))/dx)$

$lim_(xtooo)(6x)/(2x-7)$

Because the above, also, yields the indeterminate form $oo/oo$ , we can apply the rule, again:

$lim_(xtooo)((d(6x))/dx)/((d(2x-7))/dx)$

$lim_(xtooo)6/2 = 3$

L'Hôpital's rule states that the original limit goes to the same value:

$lim_(x to oo)(3x^2+3)/(x^2-7x+9) = 3$