When trying to find the slant asymptote of $\frac{2x^2+x}{x-3}$, the way I thought was correct is to divide everything by $x$ to get $\frac{2x+1}{1-\frac{3}x}$. All that was left was to say that as $x$ tends to $\infty$, $\frac{3}x$ tends to $0$, so the asymptote is $2x+1$. Spoiler: it was not. If I would do it the polynomial division way, I would get that the asymptote is $2x+7$. My question is: what is wrong with my way?
Helpful link though it did not answer my question here
On
Perhaps like this:
$(2x+1)(1-3/x)^{-1};$
Now for large $x$ expand the second factor:
$(2x+1)\cdot$
$(1 +(-1)(-3/x)+O(1/x^2))=$
$2x+1+(6x)/x+O(1/x)=$
$2x+7+O(1/x);$ and we are done.
Now try to answer where your error enters.
Used: Binomial expansion
An option:
$\dfrac{2x^2+x}{x-3}=$
$\dfrac{2((x-3)+3)^2+(x-3)+3}{x-3}=$
$2(x-3)+12+\dfrac{18} {x-3} +1+\dfrac{3}{x-3} =$
$2(x-3)+13+\dfrac{21}{x-3} =$
$2x+7+\dfrac{21}{x-3} .$
Can you finish?
On
Other people have explained how you got the wrong answer, but another way that you can use your calculation as a step to a correct answer is to recognize that this method will give you the correct slope $ 2 $ but will not tell you the constant term. So to find that, you take $$ \lim _ { x \to \infty } \Big ( \frac { 2 x ^ 2 + x } { x - 3 } - 2 x \Big ) \text , $$ and this will give you $ 7 $. Therefore, the asymptote is $ 2 x + 7 $.
Asserting that $2x+1$ is an asymptote means that$$\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(2x+1)=0.$$But$$\frac{2x^2+x}{x-3}-(2x+1)=\frac{6x+3}{x-3}$$and therefore$$\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(2x+1)=6\ne0.$$On the other hand\begin{align}\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(ax+b)=0&\iff\lim_{x\to\infty}\frac{-a x^2+2 x^2+3 a x-b x+x+3 b}{x-3}=0\\&\iff\left\{\begin{array}{l}-a+2=0\\3a-b+1=0\end{array}\right.\\&\iff\left\{\begin{array}{l}a=2\\b=7.\end{array}\right.\end{align}