\displaystyle\lim\limits_{x \to 0} \frac{\sqrt{\cos(6x)}e^{5x^2}-1}{\mathrm{tg}^2(17x)}=\\
=\lim\limits_{x \to 0} \frac{\sqrt{\cos(6x)}\cdot \frac{e^{5x^2}-1}{5x^2}\cdot 5 - \frac{1-\cos(6x)}{(6x)^2}\cdot\frac{1}{1+\sqrt{\cos(6x)}}\cdot 36} {\frac{\mathrm{tg}^2(17x)}{(17x)^2}\cdot 289}=\\
=\frac{1\cdot 1\cdot 5-\frac{1}{2}\cdot \frac{1}{2}\cdot 36}{1^2\cdot 289}=-\frac{4}{289}