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प्रश्न
Show that the function f(x) = `{(x^2, x<=1),(1/2, x>1):}` is continuous at x = 1 but not differentiable.
उत्तर
Continuity at x = 1
`(x = 1)= x^2 = (1)^2 = 1`
`lim_(x->1^+) f(x) = lim_(x->1^+) 1/x = 1`
`lim_(x->1^-) f(x) = lim_(x->1^-) x^2 = 1`
`:. f(x = 1) = lim_(x->1^-) f(x) = lim_(x->1^+) f(x) = 1`
:. f(x) is continuous at x = 1
Now differentiable at x = 1
(R.H.D at x = 1) = `lim_(x->1^+) (f(x) - f(1))/(x -1)`
`=lim_(x->1) (1/x- 1)/(x - 1)`
`= lim_(x-> 1) (-(x-1) 1/x)/(x-1)`
`= = -1/1 = -1`
(L.H.D at x = 1) = `lim_(x -> 1^(-)) (f(x)-f(1))/(x-1)`
`= lim_(x->1^1) (x^2 -1)/(x - 1) = 2`
`:. l.H.D != R.H.D`
:. f(x) is not differentiable at x = 1
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