Why L'Hospital Rule cannot apply in calculating $\lim_{x\rightarrow \infty}{(\cos {\frac{1}{x}} + \sin {\frac{1}{x^2}})^{\frac{1}{x^2}}}$ Determine the limit of the following:
$$\lim_{x\rightarrow \infty}{(\cos {\frac{1}{x}} + \sin {\frac{1}{x^2}})^{x^2}}$$
I use the logarithm of exponential function to evaluate the above:
Let $u=\frac 1 x$, then
$$\lim_{u\rightarrow 0}{(\cos {\frac{1}{x}} + \sin {\frac{1}{x^2}})^{x^2}}=\lim_{u\rightarrow 0}\exp(\frac{\ln(\cos u+\sin u^2) }{u^2})$$
Since the above expression is an indeterminate form of $\frac 0 0$, I use L'Hospital Rule to evaluate, and obtain the following result
$$-\frac{1}{2u (\sin u^2+\cos u)(\sin u-2u\cdot \cos u^2)}\rightarrow \frac 10$$ which is not true. What is my error when using L'Hospital Rule? Or there are some way to obtain the result above without using L'Hospital rule?

$$\lim_{u\to0}\dfrac{\ln(\cos u+\sin u^2)}{u^2}\rightarrow\lim_{u\to0}\dfrac{2u\cos u^2-\sin u}{2u(\cos u+\sin u^2)}$$
$$=\lim_{u\to0}\dfrac{2\cos u^2-\dfrac{\sin u}u}{2(\cos u+\sin u^2)}=\dfrac{2-1}{2(1+0)}$$
