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2015-09-15T19:55:04+05:30

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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