Why L'Hospital Rule cannot apply in calculating $\lim_{x\rightarrow \infty}{(\cos {\frac{1}{x}} + \sin {\frac{1}{x^2}})^{\frac{1}{x^2}}}$

Question

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?

Answer 1

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

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

Answer 2

Using the rules of L'Hospital we get $$e^{\lim_{x \to \infty}1/2\,{\frac {\sin \left( {x}^{-1} \right) x-2\,\cos \left( {x}^{-2} \right) }{{x}^{4} \left( \cos \left( {x}^{-1} \right) +\sin \left( {x }^{-2} \right) \right) }}}=e^0=1$$

Answer 3

You want to evaluate $$ \lim_{u\to0}\frac{\ln(\cos(u)+\sin(u^2))}{u^2} $$ The derivative of the numerator is $$ \frac{-\sin(u)+2u\cos(u^2)}{\cos(u)+\sin(u^2)} $$ because the derivative of $\ln(f(x))$ is $f'(x)/f(x)$ and not $1/(f'(x)f(x))$ as you wrote instead.

It's easier with a Taylor expansion: $$ \cos(u)+\sin(u^2)=1-\frac{u^2}{2}+u^2+o(u^2)=1+\frac{u^2}{2}+o(u^2) $$ and therefore $$ \lim_{u\to0}\frac{\ln(\cos(u)+\sin(u^2))}{u^2}= \lim_{u\to0}\frac{\ln(1+u^2/2+o(u^2))}{u^2}= \lim_{u\to0}\frac{u^2/2+o(u^2)}{u^2}=\frac{1}{2} $$