Determine the value of the constant `k` so that the function `f(x)={(sin2x)/(5x),\ \ \ if\ x!=0k ,\ \ \ if\ x=0` is continuous at `x=0` .

Explanation

Given,


`f(x)frac{sin 2 x}{5 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}{5 x}=k`



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