The new thinkings of subbu in limits [part-1]

$\displaystyle\lim_{x\rightarrow 0} sin x =\displaystyle\lim_{x\rightarrow 0} \frac{sin x}{x}\times x$ $=\displaystyle\lim_{x\rightarrow 0} 1\times 0 [since,\displaystyle\lim_{x\rightarrow 0} sin x =1]$ $=0$

Just substitute x=0 in sin(x) therefore, sin(0) =0

a basic problem. one can just substitute x=0 to get the limit as the function is not indeterminate at x=0 or discontinous(which invariably means the former)

As $sin(x)$ is continuous near $x=0$ (because it is defined at $x=0$ and its vicinity) , we have $\displaystyle\lim_{x\rightarrow 0} sin x = sin(0) = 1$ .