If $\lim _{x \rightarrow \infty}\left(1+\dfrac{a}{x}+\dfrac{b}{x^2}\right)^{2 x}=e^2$, then

The given limit is of the form $1^{\infty}$. We use the formula $\lim_{x \rightarrow \infty} (f(x))^{g(x)} = e^{\lim_{x \rightarrow \infty} g(x)(f(x)-1)}$.

Here, $f(x) = 1 + \dfrac{a}{x} + \dfrac{b}{x^2}$ and $g(x) = 2x$.

The limit becomes $e^{\lim_{x \rightarrow \infty} 2x \left(1 + \dfrac{a}{x} + \dfrac{b}{x^2} - 1\right)} = e^{\lim_{x \rightarrow \infty} 2x \left(\dfrac{a}{x} + \dfrac{b}{x^2}\right)}$.

Simplifying the expression inside the exponent:

$\lim_{x \rightarrow \infty} 2x \left(\dfrac{ax + b}{x^2}\right) = \lim_{x \rightarrow \infty} \dfrac{2ax^2 + 2bx}{x^2} = \lim_{x \rightarrow \infty} \left(2a + \dfrac{2b}{x}\right) = 2a$.

We are given that the limit is $e^2$, so $e^{2a} = e^2$.

Equating the exponents, $2a = 2$, which gives $a = 1$.

The value of $b$ does not affect the limit as $\lim_{x \rightarrow \infty} \dfrac{2b}{x} = 0$ for any finite $b$.

Therefore, $a = 1$ and $b$ can be any real number.

Answer: $a=1, b \in R$