Limit involving $(\cos x)^{1/x^4}$ I am having trouble calculating the following limit.
$$\lim_{x \to 0}(\cos x)^{1/x^4}$$
In Problems in mathematical analysis by Demidovich there is a hint that in case of $1^{\infty}$ indeterminate symbol in certain limit, one can add a term $a(x)$ which for given limit approaches zero and then after manipulation with exponents, it's easy to obtain a result of $e^{p}$, $p \in \mathbb{R}$. 
My computations in this way lead to the result of $e^{0} = 1$, but WolframAlpha says this limit equals $0$. 
If anyone could give me at least hints or solution for this problem, I would be very grateful. 
 A: Hint: 
Rewrite the term in question as
$$\cos(x)^{\frac{1}{x^4}}=e^{x^{-4}\log(\cos(x))}$$
Now use that $\cos(x)=1-\frac{x^2}{2}+O(x^4)$ as $x$ approaches $0$ and look at the Taylor series of $\log(1+x)$ at $0$.
A: Hints:
$$\lim_{x\to 0}\frac{\log\cos x}{x^4}\stackrel{\text{l'Hospital}}=\lim_{x\to 0}\frac{-\sin x}{4x^3\cos x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{-\cos x}{12x^2\cos x-4x^3\sin x}=-\infty$$
A: $$\lim_{x\to0}(\cos x)^{\dfrac1{x^4}}=\lim_{x\to0}(\cos^2x)^{\dfrac1{2x^4}}$$
$$=\left(\lim_{x\to0}(1-\sin^2x)^{\left(\frac1{-\sin^2x}\right)}\right)^{\left(-\lim_{x\to0}\dfrac{\sin^2x}{2x^4}\right)}$$
Now as $\displaystyle\lim_{h\to 0}\left(1+h\right)^{\frac1h}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e,$ the inner limit goes to $e$
Again, $\displaystyle\lim_{x\to0}\dfrac{\sin^2x}{2x^4}=\frac12\cdot\left(\lim_{x\to0}\dfrac{\sin x}x\right)^2\cdot\frac1{\lim_{x\to0 }x^2}=\frac12\cdot1\cdot\infty$
A: You received interesting answers so, I shall just report numerical results obtained using infinite precision.    
For $x=\frac{1}{10}$, the value of the logarithm of the expression is -50; for $x=\frac{1}{100}$, it is $-5,000$; for $x=\frac{1}{1000}$, it is $-500,000$; for $x=\frac{1}{10000}$, it is $-50,000,000$ and so on. I suppose you notice the beautiful factor of $100$ everytime $x$ is divided by $10$.  
On the other side, as suggested by TooOldForMath, if you develop as a Taylor series the logarithm of the expression around $x=0$, the expansion (for the logarithm) is $$-\frac{1}{2 x^2}-\frac{1}{12}-\frac{x^2}{45}+O\left(x^3\right)$$ from which you can conclude that the limit of the expression is $0$ as given by Wolfram Alpha and other answers to you post.  
Thanks for the interesting problem !
