Q.If f

True - differentiable $\Rightarrow$ continuous, converse false.

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Differentiability implies continuity, but the converse is false.

A is TRUE: Every differentiable function is continuous — this is a standard theorem.

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A piecewise function is differentiable at a point only if it is continuous there and the left and right derivatives match. For $f(x)$ at $x=0$, continuity forces $k$ to be anything, but matching derivatives forces $k=0$.

Every differentiable function is continuous.

If a function $f$ is differentiable at a point $x=a$, then

$\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=f'(a)\text{ exists (finite)}.$

Writing $f(a+h)-f(a)=\dfrac{f(a+h)-f(a)}{h}\cdot h$ and taking the limit as $h\to0$:

$\lim_{h\to0}[f(a+h)-f(a)]=f'(a)\cdot 0=0,$

which means $\displaystyle\lim_{h\to0}f(a+h)=f(a)$ — exactly the condition for continuity at $x=a$.

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Existence of a finite one-sided derivative at a point forces one-sided continuity there; since both $f'(2^+)$ and $f'(2^-)$ exist (and equal $0$), $f$ is continuous at $x=2$.

By definition, $f'(2^+) = \displaystyle\lim_{h\to 0^+}\dfrac{f(2+h)-f(2)}{h}$.

For this limit to exist and be finite (here, equal to $0$), the numerator $f(2+h)-f(2)$ must itself tend to $0$ as $h\to 0^+$ — because $f(2+h)-f(2) = h\cdot\dfrac{f(2+h)-f(2)}{h} \to 0\cdot 0 = 0$.

So $\displaystyle\lim_{h\to0^+} f(2+h) = f(2)$, i.e. $f$ is right-continuous at $x=2$.

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Differentiability at a point guarantees continuity and the existence of f′(a), but never guarantees f″(a) exists.

If $f$ is differentiable at $x=a$, then by definition the limit $f'(a) = \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}$ exists, which forces:

• (a) $f$ continuous at $a$ — always true (differentiability $\Rightarrow$ continuity).
• (b) $\lim_{x\to a} f(x)$ exists — always true, again because differentiability implies continuity.
• (c) $\lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h} = \lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$ — always true, since a two-sided derivative existing means the left-hand and right-hand derivatives are equal. …