Find the value of the constant `k` so that the given function is continuous at the indicated point: `f(x)={(x^2-25)/(x-5),\ \ \ x!=5k ,\ \ \ x=5` at `x=5`

Explanation

`L.H.L=lim_(x→5^-)(x^2−25)/(−​x−5)​`

`=lim_(x→5^-)((x−5)(x+5))/(​x−5)​`

`=lim_(x→5^−)​(x+5)`

`=5+5=10`

`R.H.L=lim_(x→5^+)​(x^2−25)/(x-5)​`

`=lim_(x→5^-)((x−5)(x+5))/(​x−5)​`

`=lim_(x→5^−)​(x+5)`

`=5+5=10`

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