$\displaystyle\lim_{x \to 0}\dfrac{(5^x - 1)^4\,\text{cosec}\,(x\log 5)}{\tan(x\log 5) \cdot \log(1 + x^2\log 25)} = \ldots\ldots$

The given limit is:

$\displaystyle\lim_{x \to 0}\dfrac{(5^x - 1)^4\,\text{cosec}\,(x\log 5)}{\tan(x\log 5) \cdot \log(1 + x^2\log 25)}$

Rewriting $\text{cosec}\,(x\log 5)$ as $\dfrac{1}{\sin(x\log 5)}$, we get:

$\displaystyle\lim_{x \to 0}\dfrac{(5^x - 1)^4}{\sin(x\log 5) \cdot \tan(x\log 5) \cdot \log(1 + x^2\log 25)}$

Dividing the numerator and the denominator by $x^4$:

$\displaystyle\lim_{x \to 0}\dfrac{\left(\dfrac{5^x - 1}{x}\right)^4}{\left(\dfrac{\sin(x\log 5)}{x}\right) \cdot \left(\dfrac{\tan(x\log 5)}{x}\right) \cdot \left(\dfrac{\log(1 + x^2\log 25)}{x^2}\right)}$

Using the standard limits $\displaystyle\lim_{x \to 0}\dfrac{a^x - 1}{x} = \log a$, $\displaystyle\lim_{x \to 0}\dfrac{\sin(kx)}{x} = k$, $\displaystyle\lim_{x \to 0}\dfrac{\tan(kx)}{x} = k$, and $\displaystyle\lim_{x \to 0}\dfrac{\log(1 + kx)}{x} = k$, we evaluate the limits of the individual terms:

Numerator limit $= (\log 5)^4$

Denominator limit $= (\log 5) \cdot (\log 5) \cdot (\log 25)$

Since $\log 25 = \log(5^2) = 2\log 5$, the denominator limit becomes:

$(\log 5) \cdot (\log 5) \cdot (2\log 5) = 2(\log 5)^3$

Substituting these back into the expression:

$\text{Limit} = \dfrac{(\log 5)^4}{2(\log 5)^3} = \dfrac{1}{2}\log 5$

Using the property of logarithms $m\log n = \log(n^m)$:

$\text{Limit} = \log(5^{1/2}) = \log\sqrt{5}$

Answer: $\log\sqrt{5}$