Prove that $\lim_{x \to 0} \frac{\sin(\sin(x))}{x} = 1$

Question

How can I prove that $\displaystyle \lim_{x \to 0} \frac{\sin(\sin(x))}{x} = 1$

Can I say that

$$\lim_{x \to 0} \sin(\sin (x)) = \lim_{x \to 0} \sin (\lim_{x \to 0} \sin (x))$$?

If that's the case, I can see the solution, because:

$$\lim_{x \to c} \sin(x) = \sin(c)$$

but I can't find a property of limits that says that:

$$\lim_{x \to 0} \sin(x) = \lim_{x \to 0} \sin ( \lim_{x \to 0} x)$$

I'm probably getting this thing all wrong.... I'm sorry if this question is too stupid. I'm learning Calculus by myself....

Thank You for your help!

Answer 1

$$\lim_{x\rightarrow0}\frac{\sin{\sin x} }{x}=\lim_{x\rightarrow0}\frac{\sin{\sin x} }{\sin x} \frac {\sin x}{x}=\lim_{\ sin x\rightarrow0}\frac{\sin{\sin x} }{\sin x} \lim_{x \rightarrow 0}\frac {\sin x}{x}=1 \cdot1$$

Answer 2

I believe you can also look at the DC term of the taylor expansion:

$$\frac{y - {y^3 \over 3!} + {y^5 \over 5!} ...}{x}\mid_{y = \sin(x)}$$

$$\frac{\left(x - {x^3 \over 3!} + {x^3 \over 5!}+...\right) - {\left(x - {x^3 \over 3!} + {x^3 \over 5!}+...\right)^3 \over 3!} + {\left(x - {x^3 \over 3!} + {x^3 \over 5!}+...\right)^5 \over 5!} ...}{x}$$

$$1 + O(x^2)$$

...which is $1$ at $x = 0$.

Answer 3

Since plugging in $x=0$ yields an indeterminant form $\dfrac{0}{0}$, apply L'Hôpital's rule to see

$$\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(\sin(x))}{x} \stackrel{\mathrm{L.H.}}{=} \displaystyle\lim_{x \rightarrow 0} \dfrac{\cos(x)\cos(\sin(x))}{1}=\cos(0)\cos(0)=1.$$

Answer 4

By definition, your limit is the derivative of $\sin(\sin(x))$ at $0$. Applying the chain rule, $\frac{d}{dx} \sin(\sin(x)) = \cos(\sin(x))\cos(x)$, so the value of the limit is $\cos(\sin(0))\cos(0) = 1$.