If f'(a) exists finitely, then show that f(x) is continuous at x = a.

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Q. If f'(a) exists finitely, then show that f(x) is continuous at x = a.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 2mImportance★★★★★

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Write $f(x)-f(a)$ as $\dfrac{f(x)-f(a)}{x-a}\cdot(x-a)$ and take the limit as $x\to a$ ; a finite derivative forces this to go to $0$ .

Since $f'(a)$ exists finitely, by definition:

$f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$

Now consider:

$\lim_{x\to a}\big[f(x)-f(a)\big] = \lim_{x\to a}\left[\frac{f(x)-f(a)}{x-a}\cdot (x-a)\right]$

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$f(2) = 4$
(a) $-2$
(b) $2$
(c) $3$
(d) $-4$
›Reveal solutionSolution

The expression is $\tfrac00$ at $x=2$ ; differentiate top and bottom (L'Hôpital) to get $f(2)-2f'(2)=-4$ .

Evaluating a $\tfrac00$ limit that encodes a derivative is a CBSE/NCERT Class 12 limits and differentiability skill.

At $x=2$ the numerator is $2f(2)-2f(2)=0$ and the denominator is $0$ , so it is $\tfrac00$ . Differentiate numerator and denominator (with $f(2)=4$ a constant): …

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CBSE 2026Set SEM31 markMCQ
$f(x) = \begin{cases} x^2 + ax + b, & x < 1 \\ x, & x \ge 1 \end{cases}$
(a) $0$
(b) $-2$
(c) $-6$
(d) $-3$
›Reveal solutionSolution

Match value and slope of the two pieces at $x=1$ : continuity gives $a+b=0$ , differentiability gives $a=-1$ ; hence $a-b=-2$ .

Continuity and differentiability of a piecewise function is a CBSE/NCERT Class 12 continuity and differentiability topic.

Continuity at $x=1$ : the left piece value equals the right piece value:

$1^2 + a\cdot1 + b = 1 \Rightarrow 1 + a + b = 1 \Rightarrow a + b = 0.$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If G(x) = -\sqrt(25-x²), then Lt(x\to1) [G(x)-G(1)]/(x-1) has the value$
(a) 1/24
(b) 1/5
(c) -√24
(d) 1/√24
›Reveal solutionSolution

This limit is exactly the definition of the derivative $G'(1)$ .

By definition, $\displaystyle\lim_{x\to 1}\frac{G(x)-G(1)}{x-1} = G'(1)$ .

Given $G(x) = -\sqrt{25-x^2} = -(25-x^2)^{1/2}$ , differentiate using the chain rule:

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Q.If f'(a) exists finitely, then show that f(x) is continuous at x = a.

Concept understanding — Differentiability

Q.If f'(a) exists finitely, then show that f(x) is continuous at x = a.

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