Have you noticed this?

Use the Trigonometric Co-Function Identities : $\tan(x) = \cot(90^\circ - x )$ . The numerator becomes $\cot \left ((10^{-n})^\circ \right )$ while the denominator becomes $\cot \left ((10^{-(n-1)})^\circ \right )$ . So the limit becomes

$\lim_{n \to \infty} \frac { \cot \left ((10^{-n})^\circ \right ) }{\cot \left ((10^{-(n-1)})^\circ \right ) } = \lim_{n \to \infty} \frac {\tan \left ((10^{-(n-1)})^\circ \right )}{\tan \left ((10^{-n})^\circ \right ) }$

Since the angles are small, we convert the angles to radians then let $\tan (x) \approx x$ .

$\frac {\tan \left [ \frac {\pi}{180} \cdot \left ((10^{-(n-1)}) \right ) \right ] } {\tan \left [ \frac {\pi}{180} \cdot \left ((10^{-n}) \right ) \right ] } \approx \frac { \left ((10^{-(n-1)})\right )}{ \left ((10^{-n})\right ) } = \boxed{10}$

I solved it similarly but used the first term of the Taylor series for $cot(x)$ . I let $x_1$ < $x_2$ and both be very small values so my limit was $\lim_{n \to \infty}\ \frac{tan(90-x_1)}{tan(90-x_2)} = \lim_{n \to \infty}\ \frac{cot(x_1)}{cot(x_2)} = \lim_{n \to \infty}\ \frac{cot(10^{-n})}{cot(10^{-(n-1)})}$

The first terms of the Taylor series for cotangent are $cot(x) = \frac{1}{x} - \frac{x}{3}-\frac{x^3}{45} ...$ so for very small $x$ the limit becomes: $\lim_{n \to \infty}\ \frac{10^{-(n-1)}}{10^{-n}}=10$ Since the final limit was a ratio of angles, it doesn't matter if we are using radians or degrees.