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$$\lim_{x\to\infty} \dfrac {x+2} {\sqrt{64x^2+1}}$$ $$\lim_{x\to-\infty} \dfrac {x+2} {\sqrt{64x^2+1}}$$ I understand how to find the limit as $x$ approaches $\infty$ but I do not understand how to find the limit when $x$ approaches $-\infty$. I entered this into Wolfram Alpha and the limit was $-\dfrac{1}{8}$. I do not understand where the negative sign came from. Is finding the limit as $x$ approaches $-\infty$ the same as finding the limit as $x$ approaches $\infty$ but adding on the negative sign at the end? Could someone explain this to me?

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    $\begingroup$ It's definitely negative, since, when $x<-2$ the function evaluates to negative, so the limit can at most be zero. $\endgroup$ – Thomas Andrews Feb 28 '13 at 23:07
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To find the limit as $x\to \infty$ carefully, one should divide top and bottom by $x$. Let us do the same thing when $x$ is negative. It turns out that we have to be careful: it is all too easy to make a mistake.

The top is uneventful, when we divide by $x$ we get $1+\frac{2}{x}$.

When we divide the bottom by $x$, it is possible to make a mistake. Note that $$\sqrt{64x^2+1}=|x|\sqrt{64+\frac{1}{x^2}},$$ (if $x\ne 0$). When we divide this by $x$, where $x$ is negative, the term $\frac{|x|}{x}$ takes on value $-1$.

The issue is that in general we cannot say that $\sqrt{y^2}=y$: this is false if $y$ is negative. But $\sqrt{y^2}=|y|$ is always true.

Remark: Negative numbers can be so treacherous that I would advise doing the problem another way. Let $u=-x$. Then for the second problem we want to find $$\lim_{u\to\infty} \frac{-u+2}{\sqrt{64u^2+1}}.$$ Now everything goes smoothly.

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  • $\begingroup$ Why is there an $|x|$ in front? I do not understand that part. $\endgroup$ – Kot Feb 28 '13 at 23:14
  • $\begingroup$ The expression $\sqrt{A}$ means the non-negative number whose square is $A$. So $\sqrt{x^2}$ is not always equal to $x$, but it is always equal to $|x|$. I suggest that you look at the remark in the post, for a less "tricky" way of doing the calculation. $\endgroup$ – André Nicolas Feb 28 '13 at 23:19
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Note that $\sqrt{x^2}=-x$ if $x<0$. Thus $ {\dfrac {x+2} {\sqrt{64x^2+1}}}={\dfrac {1+\frac{1}{x}} {\frac{\sqrt{64x^2+1}}{-\sqrt{x^2}}}}=\dfrac {1+\frac{1}{x}} {-\sqrt{64+\frac{1}{x^2}}}$ for $x<0$.

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What am about to write here is very informal but it 100% works for finding limits at $\pm \infty$ for the case polynomial/polynomial or a sqrt of a polynomial.

since we are dealing with polynomial what really controls the functoin behavior at $\pm \infty$ is the largest power of the polynomial.

in your example the highest power of the numerator is 1 so we only focus on $x$ and ignore the 2, the highest power of the denominator is also one not 2 since we have a root, so we only focus on $\sqrt{64x^2}$ and ignore the 1. but $\sqrt{64x^2}=8|x|$ and since we are taking the limit as $x\to -\infty$ , it clear that $x<0$ that is $|x|=-x$ so we have$$\displaystyle \lim_{x\to -\infty}\frac{x+2}{\sqrt{64x^2+1}}=\displaystyle \lim_{x\to -\infty}\frac{x}{\sqrt{64x^2}}=\displaystyle \lim_{x\to -\infty}\frac{x}{{8|x|}}=\displaystyle \lim_{x\to -\infty}\frac{x}{-8x}=\frac{-1}{8}.$$ again this is really informal way to deal with these kinds of limits but it works.

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In order to evaluate the limit that you have provided $$\lim_{x\to-\infty} \dfrac {x+2} {\sqrt{64x^2+1}}$$

you have to do the following steps. What's extremely important to understand is that you are finding the limit for $x \to -\infty$ meaning you are finding the limit for all numbers that are less than $0$. That is very important to understand for this problem. Let's begin.

$$\begin{align} &\tag 1\lim_{x\to-\infty} \dfrac {x+2} {\sqrt{64x^2+1}}\\ &\tag 2\lim_{x\to-\infty} \dfrac {x(1+\frac{2}{x})} {\sqrt{64x^2+1}} \\ &\tag 3\lim_{x\to-\infty} \dfrac {x(1+\frac{2}{x})} {8|x|+1} \text{Important!}\\ &\tag 4 \lim_{x\to-\infty} \dfrac {x(1+\frac{2}{x})} {-x(8+\frac{1}{x})} \\ &\tag 5 \lim_{x\to-\infty} -\dfrac {(1+\frac{2}{x})}{(8+\frac{1}{x})} \\ \end{align}$$

Alright, the first $3$ steps are fairly standard. All we are doing is factoring out the $x$ from the top and when we square the root the bottom we are left with $8|x|+1$ but since we know that $x$ has to be less than $0$, we can say instead of $|x|$ it is $-x$. That is where your negative comes from. So continuing from step $5$, we can separately take the limits of each part. The $\lim_{x \to -\infty} 8 = 8$, similarly $\lim_{x \to -\infty} 1 = 1$ and $\lim_{x\to-\infty} \frac{2}{x} = 0$ and $\lim_{x\to-\infty} \frac{1}{x} = 0$. Therefore, step $6$ looks like

$$\lim_{x\to-\infty} -\frac{(1+0)}{(8+0)} = -\frac{1}{8}$$

And there you have it. If you don't understand why $\sqrt{64x^2} = 8|x|$ then refer to this question. If you have any other questions, just comment and I will try to clarify as much as possible.

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