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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?

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    $\begingroup$ The limit does not exist because the function is not defined on a neighborhood of 3. $\endgroup$ – Did Dec 21 '18 at 16:39
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    $\begingroup$ Because the base is negative in a neighborhood of 3 hence raising it to a noninteger power cannot be done. $\endgroup$ – Did Dec 21 '18 at 16:42
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    $\begingroup$ @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. $\endgroup$ – robjohn Dec 21 '18 at 16:49
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    $\begingroup$ @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 $3$ $\endgroup$ – Shubham Johri Dec 21 '18 at 18:11
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    $\begingroup$ 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 $-1$. $\endgroup$ – Daniel Schepler Dec 21 '18 at 19:26
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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 $x \to 3$.

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  • $\begingroup$ Wait so is the problem due to the base being negative, or the power tending to 0? $\endgroup$ – Pranav Aggarwal Dec 21 '18 at 16:45
  • $\begingroup$ If I change the questions to lim x tends to 3 (x^2-5x+4)^(x-2), the answer would be -2 right? $\endgroup$ – Pranav Aggarwal Dec 21 '18 at 16:45
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    $\begingroup$ @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). $\endgroup$ – Clive Newstead Dec 21 '18 at 16:50
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    $\begingroup$ ...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. $\endgroup$ – Clive Newstead Dec 21 '18 at 16:53
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    $\begingroup$ @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. $\endgroup$ – samerivertwice Dec 23 '18 at 4:25

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