# Why does $\lim_{x \to 3} (x^2-5x+4)^{x-3}$ not exist? [closed]

Why does the following limit not exist? $$\lim_{x \to 3} \left(x^2-5x+4\right)^{x-3}$$

The base tends to $$-2$$ while the exponent tends to $$0$$. According to me the limit should be $$1$$, but the solution given for the problem says that the limit does not exist. Which is the correct answer?

• The limit does not exist because the function is not defined on a neighborhood of 3. – Did Dec 21 '18 at 16:39
• Because the base is negative in a neighborhood of 3 hence raising it to a noninteger power cannot be done. – Did Dec 21 '18 at 16:42
• @Dr.SonnhardGraubner: For any choice of the branch of log to be used, the limit would be $1$, but we don't know if we can use complex analysis or not. Certainly no mention of complex analysis was made. – robjohn Dec 21 '18 at 16:49
• @Did Technically, $3$ is a limit point of and belongs to the domain. The $\varepsilon-\delta$ definition of limit doesn't require $x\mapsto (x^2-5x+4)^{x-3}$ to be defined in the entire deleted $\delta$ neighbourhood of $3$ – Shubham Johri Dec 21 '18 at 18:11
• If you interpret the function as having domain $(-\infty, 1) \cup [(1, 4) \cap \mathbb{Z}_{(2)}] \cup [4, \infty)$ then the limit as $x\to 3$ does not exist (nor does either one-sided limit): 3 is a cluster point of the numbers $p/q$ with $p$ even and $q$ odd for which the limit approaches 1, and also a cluster point of the numbers $p/q$ with $p$ odd and $q$ odd for which the limit approaches $-1$. – Daniel Schepler Dec 21 '18 at 19:26

The quadratic $$x^2-5x+4$$ factorises as $$(x-4)(x-1)$$. If $$x$$ is in a small neighbourhood of $$3$$ then $$x-4<0$$ and $$x-1>0$$, so $$x^2-5x+4 < 0$$. This means that we run into issues with even defining the quantity $$(x^2-5x+4)^{x-3}$$ when, say, $$x = 3 \pm \dfrac{1}{2n}$$ for some positive integer $$n$$. This is bad news for the limit as $$x \to 3$$.
• @Pranav: The issue is that the base is negative, since in the limit you are required to take fractional powers in a neighbourhood of the limit. You run into the same issue if you replace the $x-3$ by $x-2$ (or $x-a$ for any $a$), so it is not the case that $\lim_{x \to 3} (x^3-5x+4)^{x-2} = -2$. You can evaluate $(x^3-5x+4)^{x-2}$ when $x=3$, and you get $-2$, but that is not the limit as $x \to 3$, since the quantity isn't even defined everywhere in an open interval about $2$ (or indeed any real number). – Clive Newstead Dec 21 '18 at 16:50
• ...this is ignoring complex numbers, of course. Taking the limit over $\mathbb{C}$ you need to consider different branch cuts of the complex plane to figure out how to define the quantity with non-integral exponents. – Clive Newstead Dec 21 '18 at 16:53
• @CliveNewstead am I right in thinking magnitude limits to $1$ although the sign is ambiguous? So the absolute value of the expression has a limit. – samerivertwice Dec 23 '18 at 4:25