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Show that f(x)=|x-3| is continuous but n...

Show that `f(x)=|x-3|` is continuous but not differentiable at `x=3` .

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The correct Answer is:
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We have, `f(x) = |x-5|`
`:. f(x) {{:(-(-x-5), if x le 5),(x-5, if x gt 5):}`
For continuity at `x = 5`.
`LHL = underset(xrarr5^(-))lim(-x+5)`
` = underset(hrarr0)(lim)[-(5-h)+5] = underset(hrarr0)(lim) h = 0`
`RHL = underset(xrarr5^(+))lim(x-5)`
`= underset(hrarr0)(lim) (5+h-5) = underset(hrarr0)(lim) h = 0`
`:. f(5) = 5-5 = 0`
`rArr LHL = RHL = f(5)`
Hence, `f(x)` is continuous at `x = 5`.
Now, `Lf'(s) = underset(xrarr5^(-))(lim)(f(x)-f(5))/(x-5)`
`= underset(xrarr5^(-))lim(-x+5-0)/(x-5)=-1`
`Rf'(5)= underset(xrarr5^(-))lim(f(x)-f(5))/(x-5)`
`= underset(xrarr5^(+))(lim)(x-5-0)/(x-5) = 1`
` :. Lf'(5) ne Rf'(5)`
So, `f(x) = |x-5|` is not differentiable at `x = 5`.
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