How can I solve this limit? $\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$

Question

$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$ Where $a$ can be any number.

I started by using $e$ like this:

$\lim _{x \to a}e^{\ln\left(\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}\right)} \to \lim _{x \to a}e^{\ln\left(\frac{\sin(x)}{\sin(a)}\right)\cdot\frac{1}{x-a}} \to \lim _{x \to a}e^{\frac{\ln\left(\sin(x)\right)-\ln\left(\sin(a)\right)}{x-a}}$

The answer should be $e^{\cot{a}}$

Answer 1

Just a little variation, but it may be nice as an alternative.

Recall that $$(1+\alpha(x))^\frac1{\alpha(x)}\to e$$ whenever $\alpha(x) \to 0$.

Replace $t= x - a$, to get \begin{eqnarray}\mathcal L &=&\lim_{t \to 0} \left[\frac{\sin(t+a)}{\sin a}\right]^\frac1t=\\ &=&\lim_{t\to 0}\left(\frac{\sin t \cos a + \cos t \sin a}{\sin a}\right)^{\frac1t}=\\ &=&\lim_{t\to 0}\left(\sin t \cot a + \cos t\right)^{\frac1t}=\\ &=&\lim_{t\to 0}\left[1+(\cos t +\sin t \cot a - 1)\right]^{\frac1t}=\\ &=&\lim_{t\to 0}\left\{\left[1+(\cos t +\sin t \cot a - 1)\right]^{\frac1{\cos t +\sin t \cot a -1}}\right\}^{\frac{\cos t + \sin t \cot a -1}{t}}=\\ &=&\lim_{t\to 0}e^{\frac{\cos t -1}t +\cot a \frac{\sin t}t}=e^{0+\cot a}=e^{\cot a}. \end{eqnarray}

Answer 2

$$\sin(x) = \sin(a) + (x-a)\cos(a) + O((x-a)^2)$$ so the argument of the limit becomes: $$(1+(x-a)\cot(a) )^{\frac{1}{x-a}}$$ Make $x-a=\epsilon$: $$(1+\epsilon\cot(a) )^{\frac{1}{\epsilon}}$$

and this is related to the number e, as you can see in: https://www.math24.net/number-e/#example3

Substitute $y=\epsilon\cot(a)$: $$(1+y )^{\frac{\cot(a)}{y}} = \left[(1+y)^{\frac{1}{y}}\right]^{\cot(a)}$$

Answer 3

You can manipulate the expression, so that

$\lim\limits_{x\to a} \left(1+\dfrac{1}{\dfrac{\sin{a}}{\sin{x}-\sin{a}}}\right)^{\dfrac{\sin{a}}{\sin{x}-\sin{a}}\dfrac{\sin{x}-\sin{a}}{\sin{a}}\dfrac{1}{x-a}}$

which is $\;\lim\limits_{x\to a}\;e^{\dfrac{\sin{x}-\sin{a}}{x-a}\dfrac{1}{\sin{a}}}=e^{\;\dfrac{\cos a}{\sin a}}$

This is how the exponent becomes $\;\cot{a}$

Edit: Maybe I skipped a few too many steps You can achieve the upper one expression by adding and subtracting one. So

$1+\dfrac{\sin{x}}{\sin{a}}-1=1+\dfrac{\sin{x}-\sin{a}}{\sin{a}}=1+\dfrac{1}{\dfrac{\sin{a}}{\sin{x}-\sin{a}}}$

The exponent is

$1\cdot \dfrac{1}{x-a}=\dfrac{\sin{a}}{\sin{x}-\sin{a}}\dfrac{\sin{x}-\sin{a}}{\sin{a}}\dfrac{1}{x-a}$

Answer 4

$$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$$

Let the limit = $L$, hence

$$L = \lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$$

Take the natural logarithm of both sides, and utilizing the fact it's continuous you get

$$ln(L) = \lim _{x \to a} \frac{1}{x-a}ln\left(\frac{\sin(x)}{\sin(a)}\right)$$

$$ln(L) = \lim _{x \to a} \frac{ln\left(\frac{\sin(x)}{\sin(a)}\right)}{x-a}$$

$$ln(L) = \lim _{x \to a} \frac{ln(sin(x))-ln(sin(a))}{x-a}$$

Now using L'hopitals rule

$$ln(L) = \lim _{x \to a} \frac{cos(x)}{sin(x)}$$

$$ln(L) = cot(a)$$

$$L = e^{cot(a)}$$

Answer 5

To make life easier, let $x=y+a$ in order to work around $y=0$. So

$$L=\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}=\lim _{y \to 0}\left(\frac{\sin(y+a)}{\sin(a)}\right)^{\frac{1}{y}}$$

$$\log(L)=\lim _{y \to 0}\frac 1y \log\left(\frac{\sin(y+a)}{\sin(a)}\right)$$ Expand the sine $$\frac{\sin(y+a)}{\sin(a)}=\frac{\sin (a) \cos (y)+\cos (a) \sin (y)}{\sin(a)}=\cot (a) \sin (y)+\cos (y)$$

A bit of Taylor series $$\cot (a) \sin (y)+\cos (y)=1+y \cot (a)-\frac{y^2}{2}+O\left(y^3\right)$$ $$\frac 1y\log\left(\frac{\sin(y+a)}{\sin(a)}\right)= \cot (a)-\frac{1}{2} \left(\cot ^2(a)+1\right)y+O\left(y^2\right)$$

Now, using $t=e^{\log(t)}$, $$\left(\frac{\sin(y+a)}{\sin(a)}\right)^{\frac{1}{y}}=e^{\cot (a)}\left( 1-\frac{1}{2} \left(\cot ^2(a)+1\right)y+O\left(y^2\right)\right)$$ which shows the limit and also how it is approached.