Is it meaningful to define the Dirac delta function as infinity at zero?

No. Defining $\delta(0)=\infty$ is not at all right, because it has to equal $\infty$ in just the "right way". You can define this as a distribution, which ignoring all technicalities, just means that it is an "evaluation linear map". I.e, given an appropriate vector space $V$ of functions $\Bbb{R}^n\to\Bbb{C}$, we consider $\delta_a:V\to\Bbb{C}$ defined as $\delta_a(f):= f(a)$. In other words, $\delta_a$ is an object which eats a function as input and spits out the value of the function a the point $a$.

Afterall, this is literally what we want it to do: when we write $\int_{\Bbb{R}^n}\delta_a(x)f(x)\,dx=\int_{\Bbb{R}^n}\delta(x-a)f(x)\,dx=f(a)$, we literally mean that $\delta_a$ is an object such that whenever we plug in a function $f$, we get out its value at $a$. Of course, writing it in this way inside an integral is a priori just nonsense, because there is no function $\delta_a:\Bbb{R}^n\to\Bbb{C}$ for which the above equality can hold true.

So, in summary, the dirac delta is a function, but it's just that the domain of the dirac delta is a space $V$ of functions, and the target space for $\delta_a$ is $\Bbb{C}$. In short, it is the "evaluation at $a$ map".

As a side remark: the dirac delta is not in any way magical/esoterical once you think of it as an evaluation map. The concept of a function as being a mapping from one set to another set is (from our luxurious perspective of having hindsight) a completely standard concept. So, in this regard the dirac delta is simply a function/mapping. The only difference is that the domain is a space of functions.

Furthermore, the concept of evaluation maps is a very basic concept in linear algebra (e.g., if you study the isomorphism between a finite-dimensional vector space and its double dual you'll see exactly what I'm talking about).

Now, the "difficulty" which comes with this is the question of "how to do calculus with these new types of objects". What I mean is in the ordinary setting of discussing functions $f:\Bbb{R}\to\Bbb{R}$, we have a notion of convergence/limits (i.e a topology), we have a notion of derivative (the study of differential calculus), and we have the notion of anti-derivatives/finding primitives etc. Extending these ideas to the more general setting is where the heart of the matter lies, and to fully appreciate that one should study more closely Laurent Schwartz's theory of distributions.

Adding to the existing answers and comments, I think a good way to argue against the slogan "$\delta(0)=\infty$" is to point out how limiting it is with respect to developing intuition for other things of the same "type" as $\delta$ itself. The $\delta$-"function" is, as others have said, not a function; rather, it's just something which can be integrated, which is not quite the same thing!

Making this rigorous leads to a lot of very interesting mathematics, to which the whole "$\delta(0)=\infty$" slogan is (in my opinion at least) a conceptual obstacle. Even if it doesn't get in the way of how one works with $\delta$ specifically, ignoring subtleties early on will just make things harder when they become central later.

The $\delta$ "function" is not really a function at all. We define $\delta$ as a sort of special notation. When we write

$\int \delta(x) f(x) dx$

This is just "syntactic sugar" (shorthand) for $f(0)$. It has many of the same properties that an ordinary function does - for example,

$\int \delta(x) (f(x) + g(x)) dx = \int \delta(x) f(x) dx + \int \delta(y) g(y) dy$

But do not mistake $\delta$ for a "real function". It is not one.

I'm not sure why it hasn't yet been pointed out that your definition of $δ$ is wrong. You certainly cannot defined it by pointwise limits as you attempted to do. Otherwise you will indeed get $δ(0) = ∞$ but it will also be completely useless and not the desired dirac-delta (e.g. you would get $2·δ(0) = ∞$ as well).

Here is another way to think of $δ$. Instead of taking limits 'prematurely', you simply refrain from taking the limit for $δ$ at all. That is, you treat $δ$ as a suitable sequence of functions, such as $(F_{1/n})_{n∈ℕ^+}$, meaning $δ$ is literally nothing more than the sequence $(F_1,F_{1/2},F_{1/3},\cdots)$. In doing so, you can then rigorously manipulate expressions involving $δ$. We first define arithmetic and integrals to apply point-wise on a sequence. For example, for sequences $p,q$ we define $p·q$ to be the point-wise product of $p$ and $q$, namely the sequence $(p_0·q_0,p_1·q_1,p_2·q_2,\cdots)$. And given sequence $r(x)$ for every $x∈ℝ$, we define $\int_ℝ r(x)\ dx$ to be $\big( \int_ℝ r(x)_0\ dx, \int_ℝ r(x)_1\ dx, \int_ℝ r(x)_2\ dx, \cdots \big)$.

Now look. For any continuous $g : ℝ→ℝ$, we have that $\int_ℝ g(x)·δ(x)\ dx$ is also a sequence, namely $\big(\int_ℝ g(x)·F_1(x)\ dx, \int_ℝ g(x)·F_{1/2}(x)\ dx, \int_ℝ g(x)·F_{1/3}(x)\ dx, \cdots \big)$ and we can easily show that this sequence tends to $g(0)$. Your $F$ works for this, but if you want further nice properties you may need smoother ones than the ones you chose (e.g. infinitely differentiable).

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[Edit: Here is my response to the later edit that quotes from a text.]

The text's definition of $δ$ is indeed completely wrong. As many people here (not just myself) have already pointed out, if you define $δ(x) = \lim_{a→0} δ_a(x)$ then definitely $δ(x) = 0$ everywhere except when $x=0$, and $δ(0) = ∞$ (using the affinely-extended reals), and definitely $\color{blue}{\int_{-∞}^∞ δ(x)\ dx = 0}$ (using the Lebesgue integral), NOT $1$, contrary to that text's erroneous claim. Note that if you do not use the affinely-extended reals, then the definition of $δ(0)$ is simply meaningless, and the whole wrong business stops there. And if you do not use the Lebesgue integral or something capable of handling isolated infinities, then you cannot even integrate $δ$. But there is no meaningful way that you can ever get $\color{red}{\int_{-∞}^∞ δ(x)\ dx = 1}$ from your text's definition, and here is why:

If $\color{red}{\int_{-∞}^∞ δ(x)\ dx = 1}$, then $\int_{-∞}^∞ δ(2x)\ dx = \int_{-∞}^∞ δ(x)\ dx = 1$ because $δ(2x) = δ(x)$ for every $x∈ℝ$ (by your text's definition of $δ$). But also $\int_{-∞}^∞ δ(2x)\ dx = \frac12 · \int_{-∞}^∞ δ(x)\ dx = \frac12$ by a simple stretch, so contradiction. This shows that if you use your text's definition of $δ$, then you do not have the desired properties for $δ$ at all.

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If, in contrast, you use the sequence-based definition I gave you, you get no such nonsense. In this section I will again use my definition. See, $\int_{-∞}^∞ δ(2x)\ dx$ is truly equal with $\frac12 · \int_{-∞}^∞ δ(x)\ dx$, because each of them is a sequence and they have identical terms! The latter is $(\frac12,\frac12,\frac12,\cdots)$, which makes complete sense.

Furthermore, if you choose your $F$ to be a sequence built from a function that is infinitely differentiable and nonzero only on a finite interval, such as a bump function (as briefly mentioned here), then you get further nice properties. For instance, $δ'$ (i.e. the term-wise derivative of $δ$) would be well-defined, and for any $g : ℝ→ℝ$ with continuous derivative and any $y∈ℝ$ we have:

  $\int_ℝ δ'(x)·g(y-x)\ dx$
  $ = \big[ δ(x)·g(y-x) \big]_{x=-∞}^∞ + \int_ℝ δ(x)·g'(y-x)\ dx$
  which converges to $g'(y)$.

To explain the last line, consider a suitable example for $δ$: Let $δ_0$ be an infinitely-differentiable function on $ℝ$ that is nonzero on $[-1,1]$ such that $\int_ℝ δ_0(x)\ dx = 1$. For each $k∈ℕ^+$ let $δ_k : ℝ→ℝ$ defined by $δ_k(x) = δ_0(x·2^k)·2^k$ for every $x∈ℝ$, so we have $δ_k$ is nonzero on $[-1/2^k,1/2^k]$ and $\int_ℝ δ_k(x)\ dx = 1$. Intuitively, $δ_k$ is nonzero on a shrinking interval as $k→∞$, but its integral over $ℝ$ remains exactly one.

Then $\big[ δ(x)·g(y-x) \big]_{x=-∞}^∞$ is a constant $0$ sequence because $\big[ δ_k(x)·g(y-x) \big]_{x=-∞}^∞ = 0$ for each $k∈ℕ$. And $\int_ℝ δ(x)·g'(y-x)\ dx$ is a sequence that converges to $g'(y)$ because:

  $\int_ℝ δ_k(x)·g'(y-x)\ dx$
  $= \int_{-1/2^k}^{1/2^k} δ_k(x)·g'(y-x)\ dx$
  $= \int_{-1}^1 δ_k(x/2^k)/2^k·g'(y-x/2^k)\ dx$
  $= \int_{-1}^1 δ_0(x)·g'(y-x/2^k)\ dx$
  $∈ \int_{-1}^1 δ_0(x)·[m_k,M_k]\ dx = [m_k,M_k]$
  where $m_k,M_k$ are the min,max of $g'(y-x/2^k)$ over all $x∈[1,1]$.

The result follows since continuity of $g'$ implies that $m_k,M_k → g'(y)$ as $k→∞$.

I actually disagree with other answers; yes, it is natural to argue that the $\delta$-function does have an infinite value at the origin. If you think of a function as a density on the real line (for example, mass density), then the "physical" definition of density would be $$ \lim_{\epsilon\to 0} \frac{m([-\epsilon,\epsilon])}{2\epsilon}, $$ the mass of the interval divided by the length of the interval. The $\delta$-function would be a density of a point mass at the origin; that is, the mass will always be $1$ and the density will be infinite.

The critiques raised in the other answers boil down to the remark that this does not tell the whole story, or that you cannot define the $\delta$-function this way. This is absolutely correct, but it is a common feature of infinity; by "replacing" something with infinity you forget some information that may or may not be relevant.

For example, if you have a sequence $a_n\to +\infty$, you may sometimes wish to "forget" the sequence, retaining only the information that $\lim a_n=+\infty$. This is very useful for solving some problems (for example, calculating $\lim (a_nb_n)$ when $b_n$ is bounded from below by a positive constant) and not useful for others (for example, calculating $\lim (a_nb_n)$ when $b_n\to 0$). The scope of usefulness basically boils down to "extending" $\mathbb{R}$ with $\pm\infty$ and partially extending the operations and relations (for example, $x+\infty=\infty$ for any $x\neq -\infty$) while leaving them undefined in some other cases (as in $\infty\cdot 0$).

So, the main problem with defining the $\delta$-function (or, rather, $\delta$-density) is the following: the main use of the density is that you can recover the mass from it by integration. If you try to do it with the $\delta$-density, you naturally run into the classical $+\infty\cdot 0$ problem (infinite density times zero length), which shows that you must have retained more information than just the infinite value an the origin. But to me, it is not per se an argument against the identity $\delta(0)=+\infty$, similarly to the fact that the $+\infty\cdot 0$ indeterminancy does not invalidate the use of infinity altogether.