# Limit of a function involving infinity

What would be the limit of $$\frac{1}{\alpha} \left(x+\frac{\alpha}{2}\right)$$ as $$\alpha$$ approaches 0, where $$-\frac{\alpha}{2} \leq x \leq \frac{\alpha}{2},\$$ and $$\alpha > 0$$?

I have tried the following, but i think it may not be correct.

$$\lim\limits_{\alpha \to 0} \frac{x}{\alpha} +\frac{1}{2}$$

Taking the limit

$$=\frac{x}{0} + \frac{1}{2} = (\infty + \frac{1}{2}) \to\infty$$

• Nitpick: to say anything is "equal to" infinity in context of typical arithmetic/calculus is at best a heavy abuse of notation and context. Infinity is moreso a "concept" than a number here. (Were it a number, then $\infty - \infty = 0$ but that would be an indeterminant limit instead.) – Eevee Trainer Mar 14 at 22:20
• What does the phrase "limit as $\alpha$ goes to zero of $\frac1\alpha(x+\frac\alpha2)$ where $-\frac\alpha2\le x\le \frac\alpha2$" even mean? – Saucy O'Path Mar 14 at 22:24
• The limit doesn't exist – Displayname Mar 14 at 22:31
• I updated my question, my apologies for abusing the notation and not being clear enough – jnxd Mar 14 at 22:45
• If $x$ is fixed then $x=0$, so the limit is $\tfrac12$. – J.G. Mar 14 at 22:48

If $$x=0$$, the limit is $$1/2$$. Suppose $$x>0$$; then $$\lim_{x\to0^-}\frac{x}{\alpha}=-\infty, \qquad \lim_{x\to0^+}\frac{x}{\alpha}=\infty$$ Similarly (with signs swapped) if $$x<0$$. Adding $$1/2$$ doesn't change the limits.

Break the problem down into cases:

Case 1: $$x=-\frac{a}{2}$$

$$\lim_{a\to0}\frac{1}{a}\left(\frac{a}{2}-\frac{a}{2}\right)=\lim_{a\to0}\frac{1}{a}(0)=0$$

This follows from $$\forall a>0.\frac{1}{a}0=0$$ for arbitrarily small $$a$$

Case 2: $$x=\frac{a}{2}$$

$$\lim_{a\to0}\frac{1}{a}\frac{2a}{2}=\lim_{a\to0}\frac{1}{a}a=1$$

This follows from $$\forall a>0.\frac{1}{a}a=1$$ for arbitrarily small $$a$$

This shows that the limit is not independent of choice of $$x$$, and therefore does not exist (this can be shown more explicitly by considering the function $$f(x,y)=\frac{1}{y}\left(x+\frac{y}{2}\right)$$. Different limits are obtained for $$(x,y)\to(0,0)$$ depending on the path you take to get to $$(0,0)$$.

While this shows that the limit does not exist, for the sake of completion, you may consider the final case.

Case 3: $$x\in\left(-\frac{a}{2},\frac{a}{2}\right)\setminus\left\{0\right\}$$

$$\lim_{a\to0}\frac{1}{a}\left(x+\frac{a}{2}\right)=\lim_{a\to0}\frac{x}{a}+\frac{1}{2}=\infty$$

Correction: As per Eevee Trainer's comment, $$\infty$$ is not a [real] number, so it is an abuse of notation to write $$\lim_{a\to0}\frac{x}{a}+\frac{1}{2}=\infty$$ assuming that $$x$$ and $$a$$ are real. However, this is exactly the case given a one-point compactification of the real number line.

In the absence of a point at infinity, if it is decided that the expression must have a limit, then that limit must be infinite. This can be proven using nonstandard techniques

$$\lim_{a\to0}\frac{1}{a}\left(x+\frac{a}{2}\right)\approx\lim_{a\to\epsilon}\frac{1}{a}\left(x+\frac{a}{2}\right)=\epsilon^{-1}\left(x+\frac{\epsilon}{2}\right)$$

Where $$\epsilon$$ is infinitesimal. It follows that $$x+\frac{\epsilon}{2}\in\text{monad}(x)$$, thus $$x\approx x+\frac{\epsilon}{2}$$. Since the reciprocal of an infinitesimal is nonfinite and $$x$$ is finite by the previous, the product of $$\epsilon^{-1}$$ and any $$y$$ s.t. $$y\in\text{monad}(x)$$ must be infinite. Therefore the limit, evaluated this way, is infinite.

STILL the limit does not exist due to the contradictions in cases 1 and 2.

You can also look at a graph of the function $$f(x,y)$$.

Edit: as per J.G.'s comment, the limit does exist for fixed $$x=0$$. This is, in fact, the only fixed value of $$x$$ for which the limit exists. However, whether or not this is "the" limit depends closely on how you evaluate the expression.

Edit: In response to KaviRamaMurthy's comment, it is worth noting that if $$x$$ is evaluated according to:

$$x=\lim_{a\to0}\left\{y\in\mathbb{R}\mid\frac{a}{2}\leq y\leq\frac{a}{2}\right\}$$

Then the result is indeterminate, whereas if $$x$$ is evaluated as a [single-valued] function of $$a$$ as in cases 1 and 2 the result is either 0, 1, or infinite.

If $$x$$ is evaluated as a constant, then the limit is either $$\infty$$ (using one-point compactification of the real-number line), $$\pm\infty$$ depending on whether $$x$$ is greater than or less than zero, or $$\frac{1}{2}$$ if $$x$$ is equal to zero.

Hopefully this covers every possible interpretation of the expression.

Yet another correction

From jnxd:

My question was transpired by the need to verify a statement $$u(x)=\lim_{\alpha\to0}u_\alpha(x)$$ mentioned in an article here. The statement is marked 4.9.

This is a different question. The statement given is $$\hspace{50pt} \delta_{\alpha}(x)=\frac{d}{dx} u_{\alpha}(x), \hspace{15pt} u(x)=\lim_{\alpha \rightarrow 0} u_{\alpha}(x) \hspace{50pt} (4.9)$$ Where $$u_\alpha(x)=\begin{cases}1&x>\frac{\alpha}{2}\\\frac{1}{\alpha}\left(x+\frac{\alpha}{2}\right)&-\frac{\alpha}{2}\leq x\leq\frac{\alpha}{2}\\0&x<\frac{\alpha}{2}\end{cases}$$ In this context, the limit refers to the function $$u(x)$$, not its value at 0. $$u(x)=\begin{cases}1&x>0\\0&x<0\end{cases}\quad\text{or}\quad u(x)=\begin{cases}1&x\geq0\\0&x<0\end{cases}$$ You will notice that smaller values of $$\alpha$$ lead to larger values of the derivative of $$u_\alpha(x)$$ in the region $$[-\frac{\alpha}{2},\frac{\alpha}{2}]$$. Because this piece of $$u_\alpha(x)$$ is linear for all $$\alpha\neq0$$, its derivative is a constant. When $$\alpha=0$$, $$u_\alpha(x)$$ becomes discontinuous at $$0$$ - it jumps from a value of $$0$$ to a value of $$1$$. The derivative of $$u(x)\vert_{x=0}$$ is thus "infinite". Ergo,

$$\delta(x)=\begin{cases}\infty&x=0\\0&\text{otherwise}\end{cases}$$

This is not the formal definition of the Dirac Delta, but it is the right idea. I think the confusion arises from the fact that the value of the Dirac Delta at $$0$$ is not well defined in terms of real numbers. The key takeaway is that the Dirac Delta can be approximated to arbitrary precision as the derivative of $$u(x)$$ - I would not consider the value of $$u(0)$$ itself.

If it is absolutely necessary to evaluate the function $$u(x)$$ at zero, then I would say $$u(0)=1/2$$. This can be determined either by evaluating the limit whilst fixing $$x=0$$ or by considering:

$$\int \delta(x)\ dx=\frac{\text{sgn}(x)+1}{2}$$

It is possible to justify this analytically (I definitely wouldn't say prove), but not without going far off topic. I would be very careful when using this, as it could easily lead to errors. Actually, if the application is probability, I would not use this at all.

• Lots of errors in your argument. Fro example, in case 3), there is no reason why $\lim \frac x a =\infty$. – Kavi Rama Murthy Mar 14 at 23:27
• @KaviRamaMurthy The reason is that the limit can be approximated to arbitrary precision using infinitesimals. Since $\epsilon^{-1}$ is not within the galaxy of $0$, it must have an absolute value larger than any real number. Hence, the limit is not finite. A much more rigorous treatment of this sort of limit is discussed in detail in Keisler's Foundations of Infinitesimal Calculus. – R. Burton Mar 14 at 23:32
• @KaviRamaMurthy Since you've said "lots of errors" would you mind clarifying the other mistakes, besides case 3, so that I may correct them? – R. Burton Mar 14 at 23:34
• $x$ is dependent on $a$ and $x \to 0$ as $a \to 0$. So the limit is indeterminate. – Kavi Rama Murthy Mar 14 at 23:34
• Statements like: 'infinity is not a number so the limit cannot be infinity' are not acceptable statements. – Kavi Rama Murthy Mar 14 at 23:37