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**Mathstud28** $\displaystyle \lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}$

To simplify things a bit we see can see a few things

Firstly you can semi-simplify things by knowing that as $\displaystyle x\to{0}$ $\displaystyle \sin(x)\sim{x}$

which gives $\displaystyle \lim_{x\to{0}}\frac{\sin(\sin(x))}{x}=\lim_{x\to{0 }}\frac{\cos(\sin(x))\cdot{\cos(x)}}{1}=1$

Or more generally $\displaystyle \lim_{x\to{0}}\frac{\sin(u(x))}{u(x)}=1,if\text{ }u(0)=0$

so from that we see that as $\displaystyle x\to{0}$ ,$\displaystyle \sin(u(x))\sim{u(x)}$

so we could just say as $\displaystyle x\to{0}$ , $\displaystyle \sin(\sin(x))\sim{\sin(x)}$

so $\displaystyle \lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}=\lim_{ x\to{0}}\frac{\sin(x)}{\sin(x)}=1$