Prove that the function `f(x) = 5x-3`is continuous at `x = 0`, at `x =-3`and at `x = 5`.

Explanation

f(x) = 5x -3
at x
Let x = - h
` rArr 0 - h to 0`
`rArr h to 0`
`f(0) = xx 0 - 3 = -3`
L.H.L. = ` underset (x to 0^(-))(lim)f(x)`
` underset (x to 0)(lim)f(0-h)= underset (x to 0)(lim) 5(-h)- 3 = 0 -3 = -3 `
`R.H.L. = underset (x to 0^(+))(lim)f(x)`
Let x = 0 + h
`rArr 0 + h to 0`
`rArr h to 0`
` underset (h to 0)(lim)f(0 + h)`
` underset (h to 0)(lim)5h - 3 = 0 -3 = -3`
`:'` L.H.L. = f(x) = R.H.L.
`:.` f(x) is continuous at x = 0 .
at x = - 3
f(-3) = 5 (-3) -3 = - 18
L.H.L. = ` underset (h to 3^(-))(lim)f(x)`
Let x = - 3 - h
`rArr - 3 h to - 3 `
`rArr h to 0 `
` underset (h to 0)(lim)f(-3 - h)`
` underset (h to 0)(lim) 5(-3 - h ) - 3 `
5(-3) - 0) - = 18
R.H.L. = ` underset (h to 0^(+))(lim)f(x)`
Let x = - 3 + h
`rArr - 3 + h to - 3`
`rArr h to 0`
` underset (h to 0)(lim)f(- 2 + h)`
` underset (h to 0)(lim)f(- 2 + h)- 3 `
`5 (-3 + 0) - 3 = - 18`
`:'` L.H.L. = f(-3) = R.H.L.
`:,` f(x) is continuous at x = - 3.
at x = 5
Let x = 5 - h
`rArr 5 - h to 5`
`rArr h to 0 `
`f(5) = 5xx5 - 3 = 22`
L.H.L = ` underset (h to 5^(-))(lim)f(x) `
`=underset(h to 0)(lim)f(5 - h)`
`underset (h to 0)(lim)5(5 - h) - 3 `
= 5 (5 - 0) - 3 = 22
R.H.L. = `underset (x to 5^(+))(lim) f(x) `
`underset (h to 0)(lim) f(5 + h ) `
`underset (h to 0)(lim) 5(5 + h )-3 `
5(5 + 0 )-3 = 22
`:'` L.H.L = F(5) = R.H.L.
`:.` F(X) IS continuous at x = 5 Hence proved.