<s>[INST] Question of computing a limit which tends to infinity $$f(x) = (x^3 + 2x^2 - 3x)/ (3x^2 + 3x - 6)$$
Question 1:
It is known that $\lim_{ x \rightarrow \infty} (f(x) - (mx +c))=0$
Find the values of $m$ and $c$,
Using the difference and sum rule and splitting out the values, simplifying $f(x)$,by dividing it with $x^2$, I would be able to prove that $f(x) = \infty$.
Same goes for $mx$ where the answer would be $\infty$ and $c$ being a constant would remain a $c$.
Hence my equation being
$\infty - \infty + c = 0$, 
Am I going about this the right way or is there anything wrong in my method of computation?
 [/INST] $\lim_{x \to \infty}((\frac {x^3+2x^2-3x}{3x^2+3x-6})-(mx+c))=0 \Rightarrow \lim_{x \to \infty} \frac {(x^3+2x^2-3x)-(mx+c)(3x^2+3x-6)}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {x^3+2x^2-3x-[3mx^3+(3m+3c)x^2+(-6m+3c)x-6c]}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {(1-3m)x^3+(2-3m-3c)x^2+(-3+6m-3c)x+6c}{3x^2+3x-6}=0.$
We want the coefficient of $x^3$ to be zero in the numerator. So $m=\frac 13$.
Now $\lim_{x \to \infty} \frac {(2-3m-3c)+\frac {(-3+6m-3c)}{(x)}+\frac {6c}{x^2}}{3+\frac 3x - \frac {6}{x^2}}=0 \Rightarrow 2-3m-3c=0 \Rightarrow 1-3c =0 \Rightarrow c=\frac 13.$
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