 large peak near $x = 0$ with area normed to $1$.
Furthermore, it is typical that the following 
function, triangular
as well, is indeed supposed to converge to the delta function - instead of becoming zero -
for $n \rightarrow \infty$ and nobody has any doubt about it.
$$
   D_n(x) = \begin{cases}
            n^2x + n & \text{if } -1/n \le x \le 0 \\
            n - n^2x & \text{if } 0 \le x \le +1/n \\
            0 & \text{everywhere else} \end{cases}
$$
The only thing that distinguishes $g_n(x)$ from $D_n(x)$ is that the maximum of
the former is shifted an infinitesimal distance $\lim_{n \rightarrow \infty}
1/(2n)$ with respect to the maximum of the latter at $x=0$.So it's easy to
see that these functions become one and the same for $n \rightarrow \infty$:
$$
  \lim_{n \rightarrow \infty} g_n(x) =
  \lim_{n \rightarrow \infty} D_n(x) = \delta(x)
$$
Therefore, in the end, we have two arguments that, unfortunately, also
lead to different outcomes for the iterated limit.

According to standard mathematics, the iterated limits do not commute:
    $$\int_0^1 \left[ \lim_{n \rightarrow \infty} g_n(x) \right] \, dx = 0
\qquad \text{and} \qquad
\lim_{n \rightarrow \infty} \left[ \int_0^1 g_n(x) \, dx \right] = 1 $$

According to this physicist, the 
iterated limits do commute:
    $$\int_0^1 \left[ \lim_{n \rightarrow \infty} g_n(x)  \right] \, dx = 1
\qquad \text{and} \qquad
\lim_{n \rightarrow \infty} \left[ \int_0^1 g_n(x) \, dx  \right] = 1 $$

And the question is, of course the following.
Is it possible to escape
large peak near $ x = 0 $ with area normed to $ 1$. 
 Furthermore, it is typical that the follow 
 affair, triangular 
 as well, is indeed supposed to converge to the delta function - rather of becoming zero - 
 for $ n \rightarrow \infty$ and nobody have any doubt about it. 
 $ $ 
    D_n(x) = \begin{cases } 
             n^2x + n & \text{if } -1 / n \le x \le 0 \\ 
             n - n^2x & \text{if } 0 \le x \le +1 / n \\ 
             0 & \text{everywhere else } \end{cases } 
 $ $ 
 The only thing that distinguishes $ g_n(x)$ from $ D_n(x)$ is that the maximum of 
 the former is shift an infinitesimal distance $ \lim_{n \rightarrow \infty } 
 1/(2n)$ with respect to the maximum of the latter at $ x=0$.So it's comfortable to 
 see that these functions become one and the same for $ n \rightarrow \infty$: 
 $ $ 
   \lim_{n \rightarrow \infty } g_n(x) = 
   \lim_{n \rightarrow \infty } D_n(x) = \delta(x) 
 $ $ 
 Therefore, in the end, we have two argument that, unfortunately, also 
 lead to different consequence for the iterated limit. 

 According to standard mathematics, the iterated limits do not permute: 
     $ $ \int_0 ^ 1 \left [ \lim_{n \rightarrow \infty } g_n(x) \right ] \, dx = 0 
 \qquad \text{and } \qquad 
 \lim_{n \rightarrow \infty } \left [ \int_0 ^ 1 g_n(x) \, dx \right ] = 1 $ $ 

 According to this physicist, the 
 iterated limits do commute: 
     $ $ \int_0 ^ 1 \left [ \lim_{n \rightarrow \infty } g_n(x)   \right ] \, dx = 1 
 \qquad \text{and } \qquad 
 \lim_{n \rightarrow \infty } \left [ \int_0 ^ 1 g_n(x) \, dx   \right ] = 1 $ $ 

 And the question is, of path the following. 
 Is it possible to escape
 lagge peak near $x = 0$ with avea normed to $1$.
Futtyermorx, it is typical that the following 
function, teiangylar
as well, is indeed rupposed no converte ti the delta functiok - inabead mh becoming zero -
for $n \rigvtarrow \infty$ dna uobody has any doubt about it.
$$
   D_n(x) = \begin{cssfs}
            g^2x + m & \tesn{in } -1/n \le x \le 0 \\
            n - n^2x & \text{if } 0 \le x \le +1/n \\
            0 & \text{everywhege epse} \end{cases}
$$
The onpy thing thqt dywtinguishes $e_n(x)$ from $D_n(x)$ is that tge maximum of
the former is shifged au infinitesumql fhstance $\lim_{i \righnarrow \infty}
1/(2n)$ with revpect tp the maximum pf vhe oatter at $x=0$.So it's easb to
see that these fonctions backme one and the sqmw for $n \rhghtxerod \ihfvy$:
$$
  \lim_{n \gigitarrow \inffy} g_n(x) =
  \lin_{n \rightarrow \infty} D_g(q) = \delta(x)
$$
Thersfore, yn the end, we have two arguments that, unfmrthnately, also
lead to difderent outcomes for tje iteratqd limit.

According to standard mathematics, the itesated uimnbf eo not commute:
    $$\int_0^1 \left[ \lim_{n \rightarrow \inftr} g_m(x) \right] \, dx = 0
\qqmad \text{and} \qquad
\lom_{j \tyghtarrow \infjy} \left[ \inf_0^1 g_n(x) \, dx \right] = 1 $$

Wccordigg to this phyficixt, the 
iterated limits do cimmute:
    $$\inn_0^1 \ledt[ \lim_{n \rightarrow \infty} g_n(x)  \rignt] \, dc = 1
\qquad \text{and} \qquad
\lnm_{n \rifhtarrow \invty} \left[ \jvt_0^1 g_n(x) \, dx  \righg] = 1 $$

Atd the question is, of courfe the foolownng.
Is it posxible eo escape
large peak near $x = 0$ with to Furthermore, it typical that the is supposed to converge the delta function instead of becoming zero - for \rightarrow \infty$ and nobody has any doubt about it. $$ D_n(x) = \begin{cases} + n & \text{if } -1/n \le x \le 0 \\ n - & } \le \le +1/n \\ 0 & \text{everywhere else} \end{cases} $$ The only thing that distinguishes $g_n(x)$ from is that the maximum of the former is an infinitesimal distance $\lim_{n \infty} 1/(2n)$ with respect to maximum the latter $x=0$.So easy see that these become one and the same for $n \rightarrow \infty$: $$ \lim_{n \rightarrow \infty} g_n(x) = \lim_{n \rightarrow D_n(x) = Therefore, in end, have arguments that, unfortunately, to different outcomes for the iterated standard mathematics, the iterated limits do not commute: \left[ \lim_{n \infty} g_n(x) \right] \, dx = \qquad \text{and} \qquad \lim_{n \rightarrow \infty} \left[ \int_0^1 \, dx \right] = 1 $$ According to this physicist, the iterated limits do commute: \lim_{n \rightarrow \infty} g_n(x) \, dx = \qquad \qquad \rightarrow \left[ \int_0^1 \, dx \right] = 1 $$ And the question is, of the following. Is it possible to escape
 large peak near $x = 0$ with area norMed to $1$.
FurthErmorE, it Is tYpIcal That The following 
fuNCtioN, triangular
as well, is indEed suPpOSed tO CoNvergE to the dELtA FUncTiOn - InsTeAD oF becoMinG zero -
foR $n \rightarrOw \iNfTy$ and nobody hAS aNy doubt aboUt iT.
$$
   D_n(x) = \begin{casEs}
            n^2X + n & \text{If } -1/N \le X \Le 0 \\
            n - n^2x & \TexT{if } 0 \le X \le +1/n \\
            0 & \teXT{everyWhere else} \EnD{Cases}
$$
THE only thING tHat dIstinguishes $g_n(x)$ frOM $D_N(X)$ is that the maxiMum of
tHe FOrMER is ShiFted an infiNiTesimAL distanCE $\lIM_{N \RigHTarrow \infty}
1/(2n)$ wIth respect tO The MaximuM oF thE Latter At $x=0$.So It'S EasY to
see that tHese Functions Become ONe and thE Same for $N \rightArrOw \iNfty$:
$$
  \LIm_{N \rIghTaRRow \INfTy} g_N(X) =
  \liM_{n \rightaRrOw \Infty} d_n(x) = \dELTA(X)
$$
TheRefOre, iN the eNd, we have two arGumEnts THat, UnforTunatEly, aLsO
lead To diffErent OuTcomes for the iteRateD limit.

AccOrdInG to StAndarD MathemAtiCs, tHe iteraTed limiTS do NoT COMmUte:
    $$\int_0^1 \left[ \lim_{n \righTaRROw \Infty} g_n(x) \Right] \, dX = 0
\QqUaD \Text{and} \qQuAd
\lIm_{n \rIGHtarrOw \inFTy} \Left[ \int_0^1 g_N(x) \, dx \riGHt] = 1 $$

acCording To This phYsIciSt, tHe 
iteRAted Limits Do commutE:
    $$\int_0^1 \lEFt[ \lim_{n \rightarrOW \infty} g_n(x)  \righT] \, Dx = 1
\QQUaD \Text{And} \Qquad
\lim_{n \riGhtaRRow \iNfty} \LEfT[ \inT_0^1 G_n(x) \, dx  \Right] = 1 $$

anD ThE Question is, of course tHe FollowIng.
Is It possible to eScape
 large peak near $x = 0$ w ith area n ormed  to  $1 $. 
Fur ther more, it is ty p ical  that the following 
f uncti on ,  tri a ng ularas well ,  i s ind ee dsup po s ed  to c onv erge to  the delta  fu nc tion - inste a dof becomin g z ero -
for $n  \r ightar ro w \ i nfty$  an d nob ody ha s  any d oubt abou ti t.
$$  D_n(x ) =\beg in{cases}
        n^2x + n & \t ext{if  } -1 / n  \l e x  \le 0 \\         n - n^ 2 x & \ t ext{if } 0 \l e x \le +1/ n  \\ 
        0 & \ text{ ev e ryw here else}\end {cases}
$ $
Theo nly thi n g thatdistin gui she s $g _ n( x) $ f ro m  $D _ n( x)$ isthat the  m ax imumof
t h e f orme r i s sh ifted  an infinites ima l di s tan ce $\ lim_{ n \r ig htarr ow \in fty}1/ (2n)$ with resp ectto the ma xim um  of  t he la t ter at  $x =0$ .So it' s easyt o
s ee t h at  these functions b ec o m eone andthe sa m efo r  $n \rig ht arr ow \ i n fty$: 
$$ \ lim_{n \ righta r ro w\infty}  g _n(x)=  \ lim _{n \ r ight arrow\infty}D_n(x )  = \delta(x)
$ $ 
Therefore, i n  t h e  e n d, w e h ave two arg umen t s th at,u nf ort u natel y, al so le a d to different outc om es for  theiterated limi t.

Accord i n g  to stan dard ma t hematics, theitera ted limits do not c ommut e:
    $ $\int_0^1 \ left[ \l im_ {n\ri ght a r ro w \infty} g_n ( x ) \r ig ht] \,dx= 0
\qq uad  \t ext {an d}  \qquad
\ lim_{n \ ri gh ta rr ow\inft y } \left[  \ int _0 ^1g_n(x )  \, dx  \rig ht]=1$ $
Accordi n gt o  thi sph ysic ist ,the iter a ted  limits  do commu te:    $$ \i nt_0^1\left[ \lim_{ n\rightarro w\in fty} g _ n (x)  \ri ght] \, dx = 1
\qquad \ t ext{and } \ qquad 
\li m_{n \rig hta rrow \ inf t y} \le ft[ \i nt_0^ 1g_n ( x ) \,d x  \ri gh t] = 1 $$A ndthe q ue stio n is, o f course the follo w ing .
Is it possi ble  toe s ca pe
 large_peak near_$x = 0$ with_area normed_to_$1$.
Furthermore, it_is_typical that the_following 
function, triangular
as_well, is indeed supposed_to converge to_the_delta function - instead of becoming zero -
for $n \rightarrow \infty$ and nobody has_any_doubt about_it.
$$
__ D_n(x) = \begin{cases}
 _    _ _   n^2x + n & \text{if_}_-1/n \le x_\le 0 \\
      _     n_- n^2x &_\text{if_}_0 \le x \le_+1/n \\
    _     _ 0 & \text{everywhere else} \end{cases}
$$
The only_thing that distinguishes $g_n(x)$ from $D_n(x)$_is that the maximum of
the_former is_shifted an infinitesimal distance $\lim_{n_\rightarrow \infty}
1/(2n)$ with_respect to_the maximum of_the latter at $x=0$.So it's easy_to
see that these_functions become one and the same_for_$n \rightarrow \infty$:
$$
__\lim_{n_\rightarrow \infty}_g_n(x) =
 _\lim_{n_\rightarrow \infty}_D_n(x)_= \delta(x)
$$
Therefore, in the end, we_have_two arguments that, unfortunately, also
lead to different_outcomes for the iterated_limit.

According_to standard mathematics, the_iterated limits do not commute:
_   $$\int_0^1 \left[ \lim_{n_\rightarrow \infty}_g_n(x) \right]_\, dx = 0
\qquad \text{and} \qquad
\lim_{n \rightarrow \infty} \left[ \int_0^1 g_n(x)_\, dx \right] = 1 $$

According_to this physicist, the_
iterated limits_do_commute:
  __$$\int_0^1 \left[_\lim_{n \rightarrow \infty} g_n(x)  \right] \,_dx =_1
\qquad \text{and} \qquad
\lim_{n \rightarrow \infty} \left[_\int_0^1 g_n(x) \, dx__\right] = 1 $$

And the question_is, of course the following.
Is it_possible to escape