What is the limit of the sine function as x approaches 1?

[Yankel]

Hello all,

I am looking for the limit of:

sin(x-1) / (x-1)

where x approaches 1. How do I do that ? Thanks !

 

[evinda]

Hello all,

I am looking for the limit of:

sin(x-1) / (x-1)

where x approaches 1. How do I do that ? Thanks !

We set $x-1=y$.
When $x \to 1$, $y \to 0$.

So, we have the limit:

$$\lim_{y \to 0} \frac{\sin y}{y}$$

and it is known that it is equal to $1$.

You can easily verify it, using L'Hôpital's rule..

 

[Fernando Revilla]

Another way, $$\lim_{t\to 0}\frac{\sin t}{t}=\lim_{t\to 0}\frac{t+o(t)}{t}=\lim_{t\to 0}\left(1+\frac{o(t)}{t}\right)=1+0=1$$ $$\underbrace{\Rightarrow}_{t=x-1} \lim_{x\to 1}\frac{\sin (x-1)}{x-1}=1$$

 

[Prove It]

We set $x-1=y$.
When $x \to 1$, $y \to 0$.

So, we have the limit:

$$\lim_{y \to 0} \frac{\sin y}{y}$$

and it is known that it is equal to $1$.

You can easily verify it, using L'Hôpital's rule...

No you most certainly can not! To use L'Hospital's Rule, you must be able to find the derivative of sin(x), so let's try it...

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \, \left[ \sin{(x)} \right] &= \lim_{h \to 0} \frac{\sin{ \left( x + h \right) } - \sin{(x)}}{h} \\ &= \lim_{h \to 0} \frac{\sin{(x)}\cos{(h)} + \cos{(x)}\sin{(h)} - \sin{(x)}}{h} \\ &= \sin{(x)}\lim_{h \to 0} \frac{\cos{(h)} - 1}{h} + \cos{(x)} \lim_{h \to 0} \frac{\sin{(h)}}{h} \end{align*}$

So to find the derivative of sin(x), we need to evaluate the very limit we are trying to find!

To find this limit, you instead have to use the geometric definition of the trigonometric functions on the unit circle.

View attachment 3663

Notice that the area of the sector is a little more than the area of the smaller triangle and a little less than the area of the bigger triangle, so when the angle x is measured in radians we have

$\displaystyle \begin{align*} \frac{\sin{(x)}\cos{(x)}}{2} \leq \frac{x}{2} &\leq \frac{\tan{(x)}}{2} \\ \frac{\sin{(x)}\cos{(x)}}{2} \leq \frac{x}{2} &\leq \frac{\sin{(x)}}{2\cos{(x)}} \\ \sin{(x)} \cos{(x)} \leq x &\leq \frac{\sin{(x)}}{\cos{(x)}}\\ \cos{(x)} \leq \frac{x}{\sin{(x)}} &\leq \frac{1}{\cos{(x)}} \\ \frac{1}{\cos{(x)}} \geq \frac{\sin{(x)}}{x} &\geq \cos{(x)} \\ \cos{(x)} \leq \frac{\sin{(x)}}{x} &\leq \frac{1}{\cos{(x)}} \end{align*}$

and now if $\displaystyle \begin{align*} x \to 0 \end{align*}$ both $\displaystyle \begin{align*} \cos{(x)} \to 1 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{\cos{(x)}} \to 1 \end{align*}$, and since $\displaystyle \begin{align*} \frac{\sin{(x)}}{x} \end{align*}$ ends up sandwiched between two ones, it must also be that $\displaystyle \begin{align*} \frac{\sin{(x)}}{x} \to 1 \end{align*}$.

Note: Technically speaking we have only proved the right hand limit, but the left hand limit is pretty much identical, just with some negatives chucked in. I'll leave that for the readers :)