Find limit: $\lim\limits_{n \rightarrow \infty} \left( \cos\frac an \right)^n$

Question

EDIT: since the proposed below "property" is incorrect in general, one can solve the limit by exponentiating the function.

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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?

Answer 1

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.

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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.

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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 (!!) $\lim_{x\to\infty} f(x)^{g(x)}$ :-)

Answer 2

What you wrote is not true. Set $f$ to be a constant function equal to $2$, and $g$ to be a constant function equal to $1$.

Then, $$\lim\limits_{x \rightarrow 0} f(x)^{g(x)} = \lim_{x\to 0} 2^1 = 2 \neq e =e^{ \lim_{x\to 0}(2-1)\cdot 1} = e^{\lim\limits_{x \rightarrow 0} (f(x)-1) \cdot g(x)}$$

However, if $f$ is strictly positive, then what is true is that.

$$\lim_{x\to\infty} f(x)^{g(x)} = e^{\lim_\limits{x\to 0} g(x)\cdot \ln(f(x))}$$

This is simply because

$$\lim_{x\to\infty} f(x)^{g(x)} = \lim_{x\to\infty} e^{\ln\left(f(x)^{g(x)}\right)} = \lim_{x\to\infty} e^{g(x)\cdot \ln(f(x))} = e^{\lim\limits_{x\to\infty} g(x)\cdot \ln(f(x))}$$

Answer 3

If $f(x)-1\to0$ then $\log(f(x))=\log(1+f(x)-1)=f(x)-1+O((f(x)-1)^2)$.

Then $f(x)^{g(x)}=\exp(g(x)\log(f(x)))=\exp(g(x)(f(x)-1)+O(g(x)(f(x)-1)^2))$.

Thus if $g(x)(f(x)-1)^2\to0$ then $\lim f(x)^{g(x)}=\lim\exp(g(x)(f(x)-1))$.

So it seems you need both $f(x)-1\to0$ and $g(x)(f(x)-1)^2\to0$ but then you're fine.

Answer 4

the property is true only if you have an indeterminate form ! If not, just pass to the limit ! And $\lim_02^1=2^1$. When there is no problem there is no problem !

A problem arise when $\lim f=1$ and $\lim g=\infty$ because $f^g=e^{g\ln f}$ and the exponent is $\infty\times0$.

If so, $\ln(f)\sim(f-1)$ (another way to say $\ln(1+x)\sim x$) and then the exponent $\sim(f-1)g$. This equivalent is an extremly powerfull tool as is $f(x)-f(a)\sim_a(x-a)f'(a)$ (just another way to read the definition of the dérivative).

I can give another example

Find $\qquad\displaystyle\lim_1\Big({\alpha+t\over\alpha+1}\Big)^{{t\over1-t}}$ !

The exponent equal $E={t\over1-t}\ln\Big({\alpha+t\over\alpha+1}\Big)$. The interior of the $\ln$ tends to 1, therefore
$\ln\Big({\alpha+t\over\alpha+1}\Big)\sim{\alpha+t\over\alpha+1}-1={t-1\over\alpha+1}$, $E\sim-{t\over\alpha+1}\sim-{\alpha\over\alpha+1}$ and the limit $\ell=e^{-{\alpha\over\alpha+1}}$

Answer 5

Solution: by using the property of continuity of exponent let's rewrite the limit as: $$\lim\limits_{n \rightarrow \infty} \left( \cos\frac an \right)^n = e^{\lim\limits_{n \rightarrow \infty}( n \cdot \ln(\cos\frac an))}$$

Then considering the power itself: $\lim\limits_{n \rightarrow \infty}( n \cdot \ln(\cos\frac an)) = \lim\limits_{n \rightarrow \infty} \dfrac{\ln(\cos\frac an)}{\frac1n} = \mbox{L'Hôpital's rule} = \lim\limits_{n \rightarrow \infty} \dfrac{-\frac{\sin \frac an}{\cos \frac an} \cdot (-a/n^2)}{- 1/n^2} = \lim\limits_{n \rightarrow \infty} - a \tan \frac an = 0$

Answer: $\lim\limits_{n \rightarrow \infty} \left( \cos\frac an \right)^n = e^{0} = 1$

Answer 6

Similarly to SmileyCraft answer, we are using that for $f(x)\to 1$ and $g(x)\to \infty$

$$\lim_{x\to x_0}f(x)^{g(x)}\stackrel{\text{identity}}=\lim_{x\to x_0}e^{g(x)\log f(x)}=\lim_{x\to x_0}e^{(f(x)-1)g(x)}$$

indeed

$$g(x)\log f(x)=g(x)\log \left(1+(f(x)-1)\right)=g(x)(f(x)-1)\frac{\log \left(1+(f(x)-1)\right)}{f(x)-1}$$

with $t=f(x)-1\to 0$ and $\frac{\log(1+t)}t \to 1$.