EDIT: since the proposed below "property" is incorrect in general, one can solve the limit by exponentiating the function.
When investigating this limit
$$\lim\limits_{n \rightarrow \infty} \left( \cos\frac an \right)^n$$
I have found a property that allows to find the limit pretty easy:
$$\lim\limits_{x \rightarrow 0} f(x)^{g(x)} = e^{\lim\limits_{x \rightarrow 0} (f(x)-1) \cdot g(x)}$$
Then a rule of derivative for $\left(\frac 00\right)$ limits is applied to get $1$ as an answer.
However, I have not been able to find any name or reference to the property above. Does anyone have a link to this property? or its name? If it is non-valid one, what's the approach to use for such a limit?
Best Answer
I couldn't immediately find this with Approach0, so I promote the comment to an answer. Will delete, when a proper duplicate target is found.
The OP (or their teacher) apparently is wont for using the rule $$ \lim_{x\to a}f(x)^{g(x)}=e^{\lim_{x\to a}(f(x)-1)g(x)}, $$ but forgot to include all the relevant assumptions. Namely that this rule is designed to ONLY handle indeterminate limits of the type $1^\infty$.
The justification for that formula comes from the well-known limits $$ \lim_{x\to\pm\infty}(1+\frac1x)^x=e. $$ A crude argument is thus to write $f(x)=1+h(x)$, when $h(x)\to0$, and $$ f(x)^{g(x)}=(1+h(x))^{g(x)}=\left((1+h(x))^{1/h(x)}\right)^{h(x)g(x)}. $$ Under the given assumptions $|1/h(x)|\to\infty$, so the base of the formula on the r.h.s. $\to e$, and the claim follows from the continuity of the exponential function.
My personal memory aid is, indeed, that of $h(x)=1/x$, $g(x)=x$.
My personal favorite application of this rule is the following proof for the limit $$ \lim_{x\to\infty}(2^{1/x}-1)x=\ln 2. $$ This is seen by the choices $f(x)=2^{1/x}\to1$ and $g(x)=x\to\infty$ when obviously $f(x)^{g(x)}=2\to2$. This time we use the formula to find the limit $$\lim_{x\to\infty} (f(x)-1)g(x)$$ using the known (!!) $\lim_{x\to\infty} f(x)^{g(x)}$ :-)