Limit of tan(x)/x as x approaching zero

[terryds]

The hint I found in http://math.stackexchange.com/quest...-lim-limits-x-to0-frac-tan-xx-1#answer-448210

limx→0(tan(x) / x)= limx→0( (tan(x)−0) / (x−0)) = limx→0 ( (tan(x)−tan(0) ) / (x−0) )=⋯

Then, I don't know how to continue it..
What identity is used ?
I don't see any similar identity to it

 

[Svein]

Hints: $tan(x)=\frac{sin(x)}{cos(x)}$ and cos(0) = 1.

 

[Mariuszek]

Furthermore use sqeeze theorem to calculate $$\lim_{x\to 0}{\frac{\sin{x}}{x}}$$
L'Hospital rule will not work in that limit

 

[Svein]

L'Hospital rule will not work in that limit

Why not?

 

[terryds]

Hints: $tan(x)=\frac{sin(x)}{cos(x)}$ and cos(0) = 1.

I know that I can make it lim x->0 tan(x) / x = lim x->0 1/cos(x) * sin(x)/x = 1 * 1 = 1

But, the hint in http://math.stackexchange.com/quest...-lim-limits-x-to0-frac-tan-xx-1#answer-448210 says

lim x->0 tan(x)/x = lim x->0 tan(x)-tan(0)/ x-0
If I plug x=0, the denominator will be zero.
Then, how to solve it using that hint ?
What trigonometric identity should be used ?

 

[Svein]

What trigonometric identity should be used ?

Well, using l'Hôpital directly, you get...

 

[Mark44]

Furthermore use sqeeze theorem to calculate $$\lim_{x\to 0}{\frac{\sin{x}}{x}}$$
L'Hospital rule will not work in that limit

Sure it will.
$\lim_{x \to 0}\frac{sin(x)}{x} = \lim_{x \to 0}\frac{cos(x)}{1} = 1$

 

[pasmith]

Sure it will.
$\lim_{x \to 0}\frac{sin(x)}{x} = \lim_{x \to 0}\frac{cos(x)}{1} = 1$

L'Hopital's rule does indeed work here, with the caveat that the argument is circular if you are trying to prove that $\sin'(0) = \lim_{x \to 0} \frac{\sin x}x$ exists and is equal to 1.