$\lim _{x \rightarrow \infty}\left(\frac{x+8}{x+1}\right)^{x+5}=\ldots$

To evaluate the limit $\lim_{x \to \infty} \left(\frac{x+8}{x+1}\right)^{x+5}$, we identify that as $x \to \infty$, the base $\frac{x+8}{x+1} \to 1$ and the exponent $x+5 \to \infty$, producing the indeterminate form $1^{\infty}$.

For limits of this form, a standard approach involves rewriting in exponential form using the identity $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x)-1]}$ when the limit exists.

Here, $f(x) = \frac{x+8}{x+1}$ and $g(x) = x+5$. First compute $f(x)-1 = \frac{x+8}{x+1} - 1 = \frac{7}{x+1}$.

Then evaluate $\lim_{x \to \infty} (x+5) \cdot \frac{7}{x+1} = \lim_{x \to \infty} \frac{7x + 35}{x+1}$. Dividing numerator and denominator by $x$ yields $\lim_{x \to \infty} \frac{7 + \frac{35}{x}}{1 + \frac{1}{x}} = 7$.

Thus, the original limit equals $e^7$.

Final answer: $\boxed{\text{D}}$