The Brainliest Answer!

This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
In the question typed above  in the denominator you have 12 x³.  i hope that is correct..

Given equation is :

 \lim_{x \to \infty} \ \frac{11x^3-3x+4}{12x^3-2x^2-7}\\

In this case, as x tends to infinity, and we take the ratio of the terms with the highest exponents of x,  from the numerator and denominator,

so :  \lim_{x \to \infty} \frac{11x^3}{12x^3}=\frac{11}{12}

Otherwise in general we can also do this way:
Let x = 1/y,    as x tends to infinity,  y tends to 0.

 \lim_{y \to 0} \frac{11/y^3-3/y+4}{12/y^3-2/y^2-7}\\\\= \lim_{y \to 0} \frac{11-3y^2+4y^3}{12-2y-7y^3}\\\\=\frac{11-0+0}{12-0-0}=11/12

3 3 3
that my mistake
its 13
do not ask me to solve any more questions... sorry..
sir really sooooo sorry
sir this question from my text . i copied from my frnd so its my mistake sooo sorry sir