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Am I allowed to use that formula with $$e$$ when you have $$1^\infty$$ in this expression : $$ \lim_{x\to a}\left(\frac{\sin(x)}{\sin(a)}\right)^\frac{1}{x-a}.$$

Edit : Some of you guys thought of completely other formula. The formula I really meant was :$$\lim_{x\to a} f^{g} = e^{\lim_{x\to a}{(f-1)g}} $$

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    Apologizing for any misunderstandings regarding this question . Im still new with this kind of formatting and stuff... – 2017-02-12
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    "that formula with $e$"? – 2017-02-12
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    Do you mean to rewrite the fraction as $\exp \ln \text{fraction}$? – 2017-02-12
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    uhmm... e^(lim[x->c] (f(x)-1)*g(x) ) assuming g(x) is potency and f(x) is the base – 2017-02-12
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    It looked like a trouble typing that formula in wanted format – 2017-02-12
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    Type that in `$$` – 2017-02-12
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    $$\lim_{x\to a} f^{g} = e^{\lim_{x\to a}{(f-1)g}} $$ : THIs is what i meant – 2017-02-12
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    @JamesGroon: this last identity makes no sense, fix it. – 2017-02-12
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    This is the exact how we were taught . I only left out the parts if lim [x→a] of f = 1 and lim [x→a] of g =$$\infty$$ – 2017-02-12
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    @JamesGroon Edited my answer to address your (clarified) question. But realize that there are literally hundreds of formulae involving $e$, and the one you were thinking of is by far not the most well-known: try to be clearer in the future. – 2017-02-12
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    I really thought is well-known "trick" . – 2017-02-12
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    @JamesGroon There are *many* "tricks" involving limits and $\exp$. Also, I am rather cautious myself, and for many I'd rather reprove them in a few lines from a core of standard tools each time -- instead of having to ponder whether the assumptions hold, and wonder what those assumptions were exactly. (In any case, upvoted the question, now that it's easier to interpret). – 2017-02-12

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With $t=x+a$, $$\left(\frac{\sin(t+a)}{\sin a}\right)^{1/t}=\left(\tan t\cot a+1\right)^{1/t}\left(\cos t\right)^{1/t}.$$

Then with $s=\tan t$,

$$\left(s\cot a+1\right)^{1/{\arctan s}}=\left(\left(s\cot a+1\right)^{1/s}\right)^{s/\arctan s}$$

which tends to $$\color{green}{\left(e^{\cot a}\right)^1}.$$

The second factor, with $s=\sin t/2$,

$$(\cos t)^{1/t}=\left(1+2\sin^2\frac t2\right)^{1/t}=\left((1+2s^2)^{1/s^2}\right)^{s^2/2\arcsin s}$$

tends to $(e^2)^0$.

So yes, you can use "that formula with $e$".

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After the edit clarifying what you meant: (original answer, and full derivation of the limit, afterwards)

  • First, let us re-derive your "formula with $e$." You have $\lim_a f^g = e^{\lim_a g\ln f}$, by definition and continuity of $ \exp$ (as long as one of the RHS or LHS exists, both do and they are equal). Now, if $\lim_a f =1$, then $\lim_a (f-1) =0 \ln f= \ln( 1+(f-1)) = (f-1)+o(f-1)$ (at $a$, using the Landau notation "little-o"), so $$\lim_a f^g = e^{\lim_a [ g(f-1) + o(g(f-1)) ] }.$$ If $g(f-1)$ is bounded around $a$, as it is in particular the case if it converges to a finite value, then you get $$\lim_a f^g = e^{\lim_a [ g(f-1) + o(1) ] } = e^{\lim_a g(f-1)}.$$

  • Back to the specific instance you have: In your case, $g(x) = \frac{1}{x-a}$ and $f(x) = \frac{\sin x}{\sin a}$, so that $f(x)-1 = \frac{\sin x-\sin a}{\sin a}$; and $$ g(x)(f(x)-1) = \frac{\sin x-\sin a}{x-a}\frac{1}{\sin a}\xrightarrow[x\to a]{} \sin'(a)\cdot \frac{1}{\sin a} = \frac{\cos a}{\sin a} $$ by definition of the derivative of $\sin$ at $a$.

    So indeed, by the above you get your limit: $e^{\frac{\cos a}{\sin a}}$.

But again, you can only use a result if you are able to argue it applies... (see also remark in the original answer) this is a fundamental aspect of mathematical proofs -- otherwise, you are following a cookbook, not writing a proof..

Let's see. I assume that by "that formula with $e$" you mean $$ \lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x = e \tag{1} $$

The short answer is:

No. If you have any doubt, or don't know whether such a result apply, then clearly you are not "allowed" to use it. You can only use things whose application you are able to justify.

Now, to deal with your limit and see what one can do: when $x\to a$, writing $x=a+h$ with $h\to 0$ (I prefer limits around zero, as a rule) $$\begin{align} \left(\frac{\sin x}{\sin a}\right)^{\frac{1}{x-a}} &= \left(\frac{\sin(a+h)}{\sin a}\right)^{\frac{1}{h}} = \left(1+\frac{\sin(a+h)-\sin a}{\sin a}\right)^{\frac{1}{h}} \\ &= \exp \frac{1}{h} \ln\left(1+\frac{\sin(a+h)-\sin a}{\sin a}\right) \end{align}$$ Since $\sin$ is continuous, we have $\frac{\sin(a+h)-\sin a}{\sin a}\xrightarrow[h\to0]{} 0$, and therefore, using the fact that $$ \frac{\ln(1+u)}{u}\xrightarrow[u\to0]{} 1, $$ we get $$ \frac{1}{h} \ln\left(1+\frac{\sin(a+h)-\sin a}{\sin a}\right) = \frac{1}{\sin a} \frac{\sin(a+h)-\sin a}{h} \frac{\ln\left(1+\frac{\sin(a+h)-\sin a}{\sin a}\right)}{\frac{\sin(a+h)-\sin a}{\sin a}} $$ By the above, the last factor has limit $1$ when $h\to 0$; the limit of the middle factor is exactly the definition of derivative of $\sin$ at $a$, i.e. $\cos a$. So overall, $$ \frac{1}{h} \ln\left(1+\frac{\sin(a+h)-\sin a}{\sin a}\right) \xrightarrow[h\to0]{} \frac{1}{\sin a}\cdot \cos a \cdot 1 = \frac{\cos a}{\sin a} $$ and by continuity of the exponential, $$ \exp \frac{1}{h} \ln\left(1+\frac{\sin(a+h)-\sin a}{\sin a}\right) \xrightarrow[h\to0]{} \boxed{e^{\frac{\cos a}{\sin a}}}. $$

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    Note that this may not be the fastest way to compute that limit; but it's pretty generalizable, and thought it to be instructive. – 2017-02-12
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    I meant on this formula : $$\lim_{x\to a} f^{g} = e^{\lim_{x\to a}{(f-1)g}} $$ – 2017-02-12
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    Sorry for misunderstanding :c – 2017-02-12
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    You have $\lim_a f^g = e^{\lim_a g\ln f}$, by definition and continuity of $\exp$ (as long as one of the RHS or LHS exists, both do and they are equal). Now, if $\lim_a f =1$, then $\lim_a (f-1) =0$ $\ln f= \ln( 1+(f-1)) = (f-1)+o(f-1)$ (at $a$, using the Landau notation), so $$\lim_a f^g = e^{\lim_a [ g(f-1) + o(g(f-1)) ] }.$$ If $g(f-1)$ is bounded, then you get $$\lim_a f^g = e^{\lim_a [ g(f-1) + o(1) ] } = e^{\lim_a g(f-1)}.$$ But again, you can only use a result if you are able to argue it applies... this is a fundamental aspect of mathematical proofs. – 2017-02-12
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Although $1^\infty$ is an indeterminate form, we can evaluate your limit. Assuming your primes are a misprint and the exponent applies to the whole fraction, we want $$\lim_{x\to a}\left(\dfrac{\sin x}{\sin a}\right)^\tfrac{1}{x-a}=\exp\lim_{x\to a}\left(\dfrac{\ln\sin x-\ln\sin a}{x-a}\right).$$By the definition of derivatives this is $\exp f'\left(a\right)$ where $f\left( x\right):=\ln\sin x$ so $f'=\cot x$, giving the result $e^{\cot a}$.

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    Altough i didn't get this derivative thing cause we havent still studied it .. this is sort of same thing using this :$$\lim_{x\to a} f^{g} = e^{\lim_{x\to a}{(f-1)g}} $$ – 2017-02-12
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$$\lim_{x\to a}\left(\dfrac{\sin x}{\sin a}\right)^{\dfrac1{x-a}}$$

$$=\left[\lim_{x\to a}\left(1+\dfrac{\sin x-\sin a}{\sin a}\right)^{\dfrac{\sin a}{\sin x-\sin a}}\right]^{\lim_{x\to a}\dfrac{\sin x-\sin a}{(x-a)\sin a}}$$

The inner limit converges to $e$

$\lim_{x\to a}\dfrac{\sin x-\sin a}{(x-a)\sin a}=\dfrac1{2\sin a}\lim_{x\to a}\dfrac{2\sin\dfrac{x-a}2\cos\dfrac{x+a}2}{\dfrac{x-a}2}$