Here's a question from the calculus section of AEEE 1997:
$${\lim_{x\to2}}\;(x-4)^x$$
I saw this question in a very old paper of the AEEE exam ($90's$ edition of the JEE exam). So when I saw this question, I obviously, wrote $4$ as the answer.
Now the answerkey said Limit does not exist.
I believed that maybe it has something to do with left hand and right hand limit and I tried referring some books on limits and understood how RHL and LHL should exist to say limit should exist but all those books have demonstrated this concept with problems involving greatest integer function or cases where limit tends to 0 which is pretty simple and understandable.
But I cant seem to apply that concept here as x approaches 2 from Right hand and Left hand side which is all positive.
So could anybody explain me the concept behind this. (And I would appreciate graphical representations!!! although not necessary) Also not I'm a high school student and starting calculus so please try to make the answer in simple language.
EDIT 1:
After reading some links from mathematics SE, I understood a possible reason for this could be 'If exponentiation $x^y$ of real numbers $x$ and $y$ is defined as $e^{y \log x}$, then one cannot define things like $(-2)^4$'
So that creates whole new level of confusion now for me. In a cubic equation say $x^3+….$, if one of the roots of the solution is -2, then…how can $x=-2$ if $(-2)^3$ isn't possible using the above logic?
EDIT 2 (Very important)
For anyone answering, please also illustrate if or why $(x+4)^x$
should exist.
Best Answer
It is true that at $x=2$ we have that $(x-4)^x=(-2)^2=4$ but recall that in the definition of limit
$$\lim_{x\rightarrow a} f(x) = L \iff \forall \varepsilon >0\, \exists \delta > 0: \forall x\in D\quad \color{red}{0<\vert x-a\vert <\delta} \implies \vert f(x)-L\vert <\varepsilon $$
we are not interested in the value of $f(x)$ at $x=a$ but we are dealing with the value of $f(x)$ in a deleted neighborhood of $x=a$.
In this case the problem is with the definition of $f(x)=(x-4)^x$ when $x-4<0 \iff x<4$ since exponentiation for negative base is not well defined.
Nevertheless, we can give a meaning to the expression and evaluate the limit for $x\in D=\left\{\frac m n\right\}$ with $m$ even integer and $n$ an odd positive integer assuming that for $a<0$ we have $a^{\frac m n}=\sqrt[n] {a^m}$. In this way limit exists and it is equal to $4$.
Refer also to the related
As noticed, for the case $\displaystyle \lim_{x\to2}\;(x+4)^x$ we have that $f(x)=(x+4)^x$ is well defined and continuous at $x=2$ therefore limit exists and is equal to $f(2)=36$.