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$$f(x) = (x^3 + 2x^2 - 3x)/ (3x^2 + 3x - 6)$$

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?

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    You are basically asked to determine the linear asymptote of $f(x)$ as $x \to \infty$, which does in fact exist since the difference between the degrees of the numerator minus denominator is $1\,$. You likely had some direction given about finding such asymptotes, so please provide more context, and where you got stuck along the way. `my equation being ∞−∞+c=0` No, that's not the way to go about it. – 2017-01-27
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    How about performing the long division $\frac{x^3 + 2x^2 - 3x}{3x^2 + 3x - 6}$ and see what you get? – 2017-01-27
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    My first question was to computer the zeros of f(x) which is basically setting f(x) as 0 and computing the value and I was asked to compute the vertical asymptotes where I set the denominator as 0 and got the answer as x tends to 2. – 2017-01-27
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    @Janedew Vertical asymtptotes do not matter, or help, here. Hint: prove first that $m = \lim_{x \to \infty} f(x)/x$. – 2017-01-27
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    You might want to cancel the common factor of the numerator and denominator first. – 2017-01-27

2 Answers 2

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For finding oblique asymptotic(s) of a curve, It's better -I think- that use function separation. Here you have $$f(x)=\frac{x^3 + 2x^2 - 3x}{3x^2 + 3x - 6}=\frac{1}{3}x+\frac13+\frac{-2x+2}{3x^2 + 3x - 6}$$ this simply shows $m=\dfrac13$, $c=\dfrac13$ and $y=\dfrac13x+\dfrac13$ is oblique asymptotic.

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$\lim_{x \to \infty}((\frac {x^3+2x^2-3x}{3x^2+3x-6})-(mx+c))=0 \Rightarrow \lim_{x \to \infty} \frac {(x^3+2x^2-3x)-(mx+c)(3x^2+3x-6)}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {x^3+2x^2-3x-[3mx^3+(3m+3c)x^2+(-6m+3c)x-6c]}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {(1-3m)x^3+(2-3m-3c)x^2+(-3+6m-3c)x+6c}{3x^2+3x-6}=0.$

We want the coefficient of $x^3$ to be zero in the numerator. So $m=\frac 13$.

Now $\lim_{x \to \infty} \frac {(2-3m-3c)+\frac {(-3+6m-3c)}{(x)}+\frac {6c}{x^2}}{3+\frac 3x - \frac {6}{x^2}}=0 \Rightarrow 2-3m-3c=0 \Rightarrow 1-3c =0 \Rightarrow c=\frac 13.$