# Thread: Limit of Trig Function

1. ## Limit of Trig Function

Find the limit of (sin x)/x as x--> 0

2. Originally Posted by magentarita
Find the limit of (sin x)/x as x--> 0
Without knowing L'Hopital's Rule, one can show this is the case using the Squeeze Theorem. You can find a proof here.

3. For simplicity
sinx can be expanded as
$\displaystyle x -\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}........$

So the Limit now becomes
$\displaystyle Lim_{x->0} \frac{x -\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}........}{x}$

$\displaystyle = Lim_{x->0} {1 -\frac{x^3}{x*3!} + \frac{x^5}{x*5!} - \frac{x^7}{x*7!}........}$
$\displaystyle = 1$

Its better if you follow Chris

4. ## Thanks but.......

Thank you but I am still a little lost here.

5. magentarita's not ready to diggest ADARSH's post since he/she's covering limits.

6. ## yes but.........

Originally Posted by Krizalid

You are right but this is the main reason why I post questions here---to learn, that it.

7. Or you can use the definition of the derivative :

$\displaystyle \lim_{x \to 0} \frac{f(a+x)-f(a)}{x}=f'(a)$

let f be the sine function, and a=0, remember that the derivative of the sine function is cosine, and observe

I guess this method suits the most what you're currently doing...

8. Originally Posted by Moo
Or you can use the definition of the derivative :

$\displaystyle \lim_{x \to 0} \frac{f(a+x)-f(a)}{x}=f'(a)$

let f be the sine function, and a=0, remember that the derivative of the sine function is cosine, and observe
How do you think the derivative of sine is derived in the first place!