I am going over my test and I am not sure what I was supposed to do for this question.
$$ \lim_{x \to -\infty} \frac {\sqrt {x^2 + 3x}}{3-2x}$$
I attempted to do the algebra but my algebra skills are too weak to do it so I resorted to logic. I stated that the top will reduce to a positive x and the bottom will reduce to a positive $2x$ giving $ \frac {x}{2x}$ which will then give me $1/2$ but this answer is wrong. Apparently the answer is correct but my reasoning is wrong.
I don't know why it is wrong but it is incredibly frustrating to me that I can get the correct answer with logical conclusions but still get the question on the test wrong. This is my second time taking calculus and at this rate I am going to fail again, no matter how hard I try I just can't get it right for some reason.
I guess this could be a broader question, but how do you take a math test? It is a mystery to me and I have absolutely no idea what a teacher expects on tests. They showed us this method in class, but on a test it is incorrect. To further complicate things I got the epsilon delta problem wrong because I didn't show the absolute value and all that stuff at the start even though I was able to show the correct epsilon.
Best Answer
Your intuition is actually correct. Most likely, your teacher wanted the 'proper' steps and justification for that intuition. To be fair to your teacher, intuition might not always be right, and without proper mathematical justification, what you write is just English :-)
We can rewrite as
$$\lim_{y \to \infty} \frac{\sqrt{y^2 - 3y}}{3 + 2y} = $$ $$\lim_{y \to \infty} \frac{y\sqrt{1 - \frac{3}{y}}}{y(\frac{3}{y} + 2)} = $$ $$\lim_{y \to \infty} \frac{\sqrt{1 - \frac{3}{y}}}{(\frac{3}{y} + 2)} = \frac{1}{2}$$
as $\frac{3}{y} \to 0$ as $y \to \infty$.