# Finding a formula for specific limits

Hey guys I have a math problem I'm not sure how to go about solving. If someone could give me a systematic method for solving these kinds of problems that would be great. The problem is as follows:

Find a formula for a function $f$ that satisfies the following conditions:

$$\lim_{x\to\pm\infty}f(x) = 0$$

$$\lim_{x\to0}f(x) = -\infty$$

$$f(2) = 0$$

$$\lim_{x\to3^-}f(x) = \infty$$

$$\lim_{x\to3^+}f(x) = -\infty$$

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@Ethan, that doesn't work because of the behaviour around $x=3$ (the function should go up on one side and down on the other), but it is helpful to plot this and see how the exponent of the $(x-3)$ term changes the behaviour. – Théophile Jul 6 '14 at 1:35
The original comment to which I responded got taken down, but the function suggested there was $\dfrac{x-2}{x^2(x-3)^2}$. Irresponsible Newb, a fun exercise that will help you better understand what's going on is to use a graphing calculator (e.g., www.desmos.com) to plot variations on these functions: for instance, what happens when you change the exponents? What difference does it make if the $(x-2)$ is in the numerator or denominator? And so on. Another post suggested $\dfrac{x-2}{x(3-x)}$, but it doesn't quite meet the conditions either. Why not? – Théophile Jul 6 '14 at 1:55

Systematic method: You have different behavior at $-\infty, 0, 2, 3, +\infty$. Hence, just build a piecewise-defined function: $$f(x)=\begin{cases} a(x) & x<0\\ b(x) & 0<x<2\\ c(x) & 2\le x< 3 \\ d(x) & 3<x\end{cases}$$

Now just find functions $a,b,c,d$ to satisfy the desired parameters.

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okay, so first, when analyzing your problem. Notice certain things:

• If you have a left and right limit at an x-coordinate that don't equal one another and go to from negative to positive infinity and vice versa, you can assume it is a vertical asymptote.
• If you have the limit as f(x) tends to infinity equal to 0, chances are the degree of the denominator is greater than the degree of the numerator. To put it in simpler terms, the horizontal asymptote is y = 0.
• Since we are dealing with fractions, the limit as x tends to 0 being negative infinity would indicate that we have a difference that is even between the degree on the top and the degree on the bottom and a negative sign out front. For more clarification look at the graphs of $f(x) = \frac{1}{x}$ and $f(x) = \frac{1}{x^2}$

So:

$$f(x) =- \frac{x-2}{x^2(x-3)}$$

• Asymptote at x=3
• $f(2) = 0$
• Degree of top is 1, degree of bottom is 3, difference is even (2).
• Degree of denominator is greater than numerator; horizontal asymptote at y = 0.
• Negative sign allocates for the limit as x tends to 0 to be negative infinity.

Take a look at the graph.

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"If you have a left and right limit at an $x$-coordinate that don't equal one another, it's most likely a vertical asymptote." No, a vertical asymptote occurs when $\lim_{x\to a^+}f(x) = \pm\infty$ or $\lim_{x\to a^-}f(x) = \pm\infty$ for some $a$. – Théophile Jul 6 '14 at 1:33
There is no reason to reply to me so rudely, Varun. I indeed read that you wrote "most likely", which is misleading and doesn't explain what's going on. If, say, $\lim_{x\to 0^-}f(x) = 3$ and $\lim_{x\to 0^+}f(x) = 4$, then it is certainly not the case that $f$ "most likely" has a vertical asymptote there. – Théophile Jul 6 '14 at 1:38
@Théophile I apologize for my rudeness, I realize that my answer can be misleading and since it asked for a detailed explanation I will change as accordingly. upvoted. – Varun Iyer Jul 6 '14 at 1:39
@Ethan I am sorry if it may look like I had stolen your answer. But also note that your answer doesn't allocate for the negative and you simply put your answer without any details or explanations as to how you arrived at that. – Varun Iyer Jul 6 '14 at 1:41
@Ethan Understand that at Math S.E., people don't try to "steal" answers. They offer hints and answers to help other people understand their mathematical questions. For the record, his answer isn't 100% yours. – Vishwa Iyer Jul 6 '14 at 1:43

Here's a general approach to find a rational function that has the behavior you're looking for. In particular, I'm assuming you've been given the following:

The values of $f$ at certain points are zero.

The limits of $f$ approaching certain points from above or below are $\pm \infty$; these points include $x \to \pm \infty$, where the limit may also be specified as $0$.

I'm going to assume that the function specified should have no other "infinities" than the ones specified, but may have other zeroes. I'll further need to assume that the limits as $x \to \pm \infty$ are identical, either both zeroes or both infinities (possibly of opposite signs), or the problem cannot be solved with rational functions. I'll mention at the end a way to handle this using exponentials, though.

1. [Pick your breakpoints] Write down the breakpoints (the ones where the limiting behavior -- either "has value 0" or "goes to $+\infty$" or "goes to $-\infty$) is specified. Call them $a_1, a_2, \ldots n_k$. In your case, these are

$$a_1 = 0\\ a_2 = 2 \\ a_3 = 3,$$ and $k = 3$, because there are 3 breakpoints, while $$b_0 = -1\\ b_1 = 1\\ b_2 = 2.5 \\ b_3 = 4.$$

We'll also need the behavior as $x \to \pm \infty$, but let's leave that for later.

For each pair of breakpoints, pick another point between them, with extra points at the ends:

\begin{align} b_0 = a_1 - 1 \\ b_1 = \frac{1}{2} (a_1 + a_2) \\ \ldots b_{k-1} = \frac{1}{2} (a_{k-1} + a_k) \\ b_{k} = a_k + 1 \end{align}

1. [Make the behavior at breakpoints look right] Write down a first function that looks like this:

$$f_1(x) = C (x-b_0)^{m_0}(x - a_1)^{n_1} (x-b_1)^{m_1}(x-a_2)^{n_2} \cdots (x-a_k)^{n_k} (x-b_k)^{m_k}$$ We're going to pick the exponents $n_i$ to be either $+1, -1,$ or $-2$, and the exponents $m_i$ to be either $0$ or $1$.

(a) If the value at $a_i$ is zero, pick $n_i = 1$.

(b) If the limit from both sides is $+\infty$, or from both sides is $-\infty$, pick $n_i = -2$.

(c) If the limits from the two sides are infinities of opposite signs, pick $n_i = -1$.

(d) Look at each interval between breakpoints. Suppose, for instance, that $a_3 = 2$ and $a_4 = 5$ are adjacent breakpoints, and you're told $$\lim_{x\to 2^{+}} = +\infty \\ \lim_{x\to 5^{-}} = -\infty$$ Then the function must change signs on the interval $[2, 5]$; in this case, we set the exponent $m_3$ for $(x - b_3)$ to be $+1$.

On the other hand, if, between breakpoints, there's no sign-change, set the exponent $m_i$ for that interval to zero.

If your function is specified to have a value $0$ at some $a_i$, you should regard it as having a sign-change there.

Let's apply this to your example:

At $-\infty$, the value is $0$. We can assume that as we approach $-\infty$, we approach zero from above, so we'll say that the value on the first interval, $-\infty < x < 0$, is positive at the left end, but becomes negative near $x = 0$ (because the limit at 0 is $-\infty$). So using rule d, we pick $m_0 = 1$.

At $x = a_1 = 0$, the limit from both sides is $-\infty$, so rule $b$ says $n_1 = -2$.

On the interval $0 < x < 2$ (i.e., from $a_1$ to $a_2$), the function must go from $-\infty$ to $0$; we'll choose to make it always negative --- that's the easiest choice --- so $m_1 = 0$. At $x = a_2 = 2$, the function has to be zero, so rule "a" says that $n_2 = 1$. And the "sign changes at a zero" rule says that just to the right of $a_2$, the function is positive.

On the interval $2 < x < 3$, i.e., $a_2 < x < a_3$, the function starts out positive and must reach $+\infty$, so we'll make it positive the whole time, i.e., pick $m_2 = 0$. The limits from opposite sides at $x = a_3 = 3$ have opposite signs, so rule "c" says that $n_2 = -1$.

To the right of $x = a_3 = 3$, the function is initially negative; when we reach $\infty$, we're told the limit must be zero, so we can choose a sign for what happens as it approaches $\infty$; let's pick it to approach from below, so that $m_3 = 0$.

We're now almost done. We have $$f_1(x) = (x- (-1)) x^{-2} (x - 2)^1 (x - 3)^{-1} = \frac{(x+1)(x-2)}{x^2 (x-3)}.$$ The only question that remains is whether the limits of $f_1$ as $x \to \pm \infty$ are correct.

1. [Adjust the limits at infinity] Let $n$ be the sum of all the exponents in $f_1$. In our case, this is $n = 1 + 1 + (-2) + (-1) = -1$.

There are now several cases.

Case 1: There's no breakpoint with an infinity assigned; in this case, only zeroes have been specified. If the limit at either end is specified as zero, then there's no rational function that will do what's needed. We can, however, still find a function:

Case 1a: one limit is zero, the other infinity: Multiply $f_1$ by either $e^x$ or $e^{-x}$ to achieve the desired behavior at $\pm \infty$.

Case 1b: both limits are zero: multiply $f_1$ by $e^{x^2}$ to get the result.

Case 2: There's a breakpoint, $a_p$, where the limit is $\pm \infty$ (or the one sided limits are both infinities, possibly of opposite signs). Let $s = n$ if $n$ is even, and $s = n+1$ if $n$ is odd. And let $r = 0$ if $n$ is even, and $1$ if $n$ is odd. $$f_2(x) = f_1(x) (x- a_p)^{-s} (x - b_0)^r.$$ This is a function of total degree zero whose limits at $\pm \infty$ are both $+1$.

Case 2a: If the desired limits are both $0$, let $$f_3(x) = f_2(x) (x - a_p)^{-2}.$$

Case 2b: If the desired limits are both $+\infty$, let $$f_3(x) = f_2(x) (x - b_0)^2.$$

Case 2c: If the desired limits are both $-\infty$, let $$f_3(x) = -f_2(x) (x - b_0)^2.$$

Case 2d: If the desired limits are $-\infty$ and $+\infty$, let $$f_3(x) = \pm f_2(x) (x - b_0).$$

Case 2e: if one desired limit is zero and the other infinite, multiply $f_2$ by either $e^x$ or $e^{-x}$, and then throw in a factor of $-1$ to adjust the sign as needed.

Applying all this to your case, by the time we've built $f_1(x)$, we already have the correct limits at $\pm \infty$. But following through the algorithm, we're told to find a breakpoint where infinity is the assigned value...let's use $a_3 = 3$. Case 2 tells us to build \begin{align} f_2(x) &= f_1(x) (x - 3)^0 (x-(-1))\\ & = \frac{(x+1)(x-2)}{x^2 (x-3)} (x-(-1)) \\ & = \frac{(x+1)^2(x-2)}{x^2 (x-3)}. \end{align}

We now fall into case 2a, since both desired limits at $\infty$ are zero. We're instructed to multiply by $(x-3)^{-2}$, so we get \begin{align} f_3(x) &=\frac{(x+1)^2(x-2)}{x^2 (x-3)^3}. \end{align}

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