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Is every differentiable function cont...

Is every differentiable function continuous?

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First part :
Let a function f be differentiable at x = a.
Then `underset(hrarr0)lim(f(a+h)-f(a))/(h)` exists and
`underset(hrarr0)lim(f(a+h)-f(a))/(h)=f'(a)" …(1)"`
In order to show that f is continuous at x = a,
We have to show that `underset(xrarra)lim` f(x) = f(a),
i.e., `underset(xrarra)lim[f(x)-f(a)]=0`
Put x = a+h. Then as `xrarra, hrarr0` and we have to show that
`underset(hrarr0)lim` [f(a+h)-f(a)]=0
Now `underset(hrarr0)lim [f(a+h)-f(a)]`
`=underset(hrarr0)lim(f(a+h)-f(a))/(h)h" "...(because h rarr 0, h ne 0]`
`=[underset(hrarr0)lim(f(a+h)-f(a))/(h)](underset(hrarr0)limh)`
`=f'(a)xx0=0" ... [By (1)]"`
`therefore` f is continuous at x = a.
Second Part :
The converse of the statement is 'Even if a function f is continuous at x = a, it need not be differentiable at x = a'.
Consider the function f(x) = |x|. It is obvious that
`underset(xrarr0^(-))limf(x)=underset(xrarr0^(+))limf(x)=f(0)=0`
`therefore` the function f is continuous at x = 0.
For `h lt 0, |h| = - h`.
`therefore underset(hrarr0^(-))lim(f(0+h)-f(0))/(h)=underset(hrarr0^(-))lim(|h|-|0|)/(h)`
`=underset(hrarr0)lim(-h)/(h)=underset(hrarr0)lim(-1)=-1" ...(1)"`
For `hgt0, |h|=h`.
`therefore underset(hrarr0^(+))lim(f(0+4)-f(0))/(h)=underset(hrarr0^(+))lim(|h|-|0|)/(h)`
`=underset(hrarr0)lim(h)/(h)=underset(hrarr0)lim1=1" ... (2) ..."[hrarr0, therefore hne0]`
`therefore` from (1) and (2),
`underset(hrarr0^(-))lim(f(0+h)-f(0))/(h)neunderset(hrarr0^(+))lim(f(0+h)-f(0))/(h)`
`therefore underset(hrarr0)lim(f(0+h)-f(0))/(h)` does not exist
`therefore` f is not differentiable at x = 0.
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