Let $$\delta_c(t)=\begin{array}{cc} \big\{ & \begin{array}{cc} 0 & c<t&t<-c \\ \frac{1}{2c} & -c\leq t\leq c \\ \end{array} \end{array}$$ Then $\lim_{c\to0^+}\delta_c(t)$ is the Dirac Delta function for $\mathbb R^2$
However if we let $L=\lim_{c\to+\infty}$ for shorthand $$L\delta_c(t)=\begin{array}{cc} \big\{ & \begin{array}{cc} 0 & L(c)<t&t<L(-c) \\ L(\frac{1}{2c}) & L[-c\leq t\leq c] \\ \end{array} \end{array}=\begin{array}{cc} \big\{ & \begin{array}{cc} 0 & L(c)<t&t<L(-c) \\ 0 & L[-c\leq t\leq c] \\ \end{array} \end{array}$$ Which in English means that $t<-\infty$, the limit is $0$, and that $t>\infty$ the limit is $0$, and since these regions are absurd, we can disregard them. However for for $-\infty\le t\le\infty$ the limit is $0$, so for all real numbers, the limit is $0$, so the limit as a whole is $0$, so $L\delta_c(t)=0$
And we know that $$\int_{-\infty}^\infty \delta_c(t)dt=\int_{-c}^c\frac1{2c}dt=1$$
But consider$$L\int_{-\infty}^\infty\delta_c(t)dt=\int_{-\infty}^\infty L\delta_c(t)dt=\int_{-\infty}^{\infty}0dt=0$$ and $$L\int_{-\infty}^\infty\delta_c(t)dt=L1=1$$
So what happened? I know that I messed up somewhere but I don't see it. I'm assuming the error is where the limit is $0$ but I don't see where I messed up unless $L\frac1{2c}\ne0$
I'm guessing by "$L=\lim_{c\to+\infty}$" you're just introducing a shorthand notation for taking this limit, like a limit operator for the pointwise limit with respect to $t$. In that case the mistake is here: $$\lim_{c\to+\infty} \int_{-\infty}^{+\infty} \delta_c(t)\,dt \color{red}{=} \int_{-\infty}^{+\infty} \lim_{c\to+\infty} \delta_c(t)\,dt.$$ (I rewrote it using the usual limit notation.)
In general, you can't switch a limit with an integral, i.e. in general $$\lim_{c\to c_0} \int_a^b f_c(t)\,dt \neq \int_a^b \lim_{c\to c_0} f_c(t)\,dt.$$ The equality is guaranteed to hold if the functions $f_c(t)$ converge uniformly to $f(t)=\lim\limits_{c\to c_0} f_c(t)$ on the interval $[a,b]$. But I suspect that the convergence of $\delta_c(t)$ to $\lim\limits_{c\to+\infty} \delta_c(t)=0$ (the identically zero function) on $t\in(-\infty,+\infty)$ is not uniform.