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Limit at Negative Infinity of a Quotient with Radicals

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Okay so I am learning the following type of limits right now... $$\lim_{x\rightarrow ∞} \frac{\sqrt{9x^2+2}}{2x-9}$$ Just Dandy. So easy, I know exactly how to solve. But then something horrible happens. A devil of a twist in such a joy of a problem. $$\lim_{x\rightarrow -∞} \frac{\sqrt{9x^2+2}}{2x-9}$$

When ever the limit is to negative infinity I always get the right numeral but my sign for the number is always wrong. Using the following process I can consistantly get the right answer of 3/2 but without knowing if it is positive or negative. An issue I don't have when the limit is to positive infinity. I am sure this has something to do with the following facts.

$$\sqrt{x^2}=|x|$$ and $$|x|=x$$ if $$x ≥ 0$$ and $$-x$$ if $$x < 0$$

So here it goes...

Step #1 Write the problem

$$\lim_{x\rightarrow -∞} \frac{\sqrt{9x^2+2}}{2x-9}$$

Step #2 Multiply the expression by the reciprocal of the degree (for limits to infinity, i.e. n=1), making sure to take the reciprocal of the square root of the square of x in the numerator so it can be simplifiied with the existing radical

$$\lim_{x\rightarrow -∞} \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{x}}* \frac{\sqrt{9x^2+2}}{2x-9}$$

Step #3 Rewrite as a complex fraction

$$\lim_{x\rightarrow -∞}\frac{\frac{\sqrt{9x^2+2}}{\sqrt{x^2}}}{\frac{2x-9}{x}}$$

Step #4 Simplify

$$\lim_{x\rightarrow -∞} \frac{\sqrt{9+\frac{2}{x^2}}}{2-\frac{9}{x}}$$

Step #5 Apply the rule where if the degree in the denominator is greater then numerator then the limit to infinity of that expression is 0

$$\frac{\sqrt{9}}{2}$$

Step #6 Simplify

$$\frac{3}{2}$$

Now that is all fine and if this was to positive infinity I would be 100% sure that positive 3 over 2 is the answer but to negative infinity I would not be able to tell you the sign. (the correct answer is -3/2 but I used a cheaty calculator to get that) It is not just always negative, sometimes it is positive anyways. How do I factor for the negative infinity in my work. I tried making the reciprocal of the degree factor with the radical in it (numerator in this problem) negative before continuing to account for the absolute value effect but that does not help either. I keep getting incosistent answers that may but not nessisarly be right. What do I need to do differently to solve problems of this kind correctly?

P.S. This took a while to write but I am in no rush for an answer. I am self-studying AP Calculus BC. Thanks in advance. :D

Oct 7, 2018

#1
+4781
+1

I suppose one trick you could use is

$$\lim \limits_{x\to -\infty}~f(x) = \lim \limits_{x \to \infty}~f(-x)$$

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Oct 7, 2018
#2
+27664
+2

You might be making this more complicated than necessary!

As $$x\rightarrow -\infty$$ the 2 in the numerator is negligible compared with 9x2, and the -9 in the denominator is negligible compared with the 2x.  So:

$$\lim_{x\rightarrow -\infty}\frac{\sqrt{9x^2+2}}{2x-9}\rightarrow\lim_{x\rightarrow -\infty}\frac{\sqrt{9x^2}}{2x}\rightarrow \lim_{x\rightarrow -\infty}\frac{3|x|}{2x}$$

Now, when x is negative $$\frac{|x|}{x}=-1$$ so the limit is -3/2.

Oct 7, 2018