For what value of `k` is the function `f(x)={(sin2x)/x ,\ \ \ x!=0k ,\ \ \ x=0` continuous at `x=0` ?

Explanation

`f(x)frac{sin 2 x}{x}, & x equiv 0 k, & x=0`



Since `{f}({x})` is continuous at `{x}=0`.


Therefore,


`lim _{x rightarrow 0} f(x)=f(0)`



`Rightarrow lim _{x rightarrow 1} frac{sin 2 x}{x}=k`



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