क्या `f(x)={{:(x",","यदि ",x le1),("5,","यदि ",xgt1):}` द्वारा परिभाषित फलन `f, x=0,x=1` तथा x = 2 पर सतत है?

Explanation

`f(x)={{:(x",","यदि ",x le1),("5,","यदि ",xgt1):}`
x = 0 पर
`L.H.L.=underset(xrarr0^(-))(lim)f(x)" माना "x=0-h`
`=underset(hrarr0)(lim)f(0-h)" "rArr 0-h rarr0`
`=underset(hrarr0)(lim)(0-h)=0" "rArr" "hrarr0`
f(0)=0
`R.H.L.=underset(xrarr0^(+))(lim)f(x)" माना "x=0+h`
`=underset(hrarr0)(lim)f(0+h)" "rArr 0+h rarr0`
`=underset(hrarr0)(lim)(0+h)=0" "rArr" "hrarr0`
`because" L.H.L. "=f(0)=" R.H.L."`
`therefore" "f(x),x=0` पर सतत है |
`f(1)=1`
`L.H.L.=underset(xrarr1^(-))(lim)f(x)" माना "x=1-h`
`=underset(hrarr0)(lim)f(1-h)" "rArr1-h rarr1`
`=underset(hrarr0)(lim)(1-h)=1-0 =1" "rArr" "hrarr0`
`R.H.L.=underset(xrarr1^(+))(lim)f(x)" माना "x=1+h`
`=underset(hrarr0)(lim)f(1+h)" "rArr1+h rarr1`
`=underset(hrarr0)(lim)5=5" "rArr" "hrarr0`
`because" "L.H.L. ne R.H.L.`
`therefore" "f(x),x = 1 ` पर सतत नहीं है।
x = 2 पर
f(2) = 5
`L.H.L. = underset(xrarr2^(-))(lim)f(x)" माना "x=2-h`
`=underset(hrarr0)(lim)f(2-h)" "rArr 2-h rarr2`
`=underset(hrarr0)(lim)5=5" "rArr" "hrarr0`
`R.H.L. = underset(xrarr2^(+))(lim)f(x)" माना "x=2+h`
`=underset(hrarr0)(lim)f(2+h)" "rArr 2+h rarr2`
`=underset(hrarr0)(lim)5=5" "rArr" "hrarr0`
`because R.H.L. = f(2) = R.H.L.`
`therefore f(x), x=2 `पर सतत है।