Check the continuity of the function $$f(x) = \begin{cases} x+2, & x 1 \end{cases}$$ at $x=1$.

To check continuity at $x=1$, all three conditions must be tested in turn. **Condition 1 — is $f(1)$ defined?** Yes: by the middle piece of the definition, $f(1)=5$. **Condition 2 — does $\lim_{x \to 1} f(x)$ exist?** Check both one-sided limits. - Left-hand limit (using $x+2$ for $x 1$): $$\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} 3x = 3(1) = 3$$ Since LHL $=$ RHL $=3$, the limit exists: $\lim_{x \to 1} f(x) = 3$. **Condition 3 — does the limit equal $f(1)$?** The limit is $3$, but $f(1)=5$. Since $3 \neq 5$, condition 3 **fails**. …

Check the continuity of the function f(x) = x+2, x<1 \5, x=1 \3x, x>1 at x=1.

Exercises · Q12

f(x) = \begin{cases} x+2, & x<1 \\ 5, & x=1 \\ 3x, & x>1 \end{cases}

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To check continuity at $x=1$ , all three conditions must be tested in turn.

Condition 1 — is $f(1)$ defined? Yes: by the middle piece of the definition, $f(1)=5$ .

Condition 2 — does $\lim_{x \to 1} f(x)$ exist? Check both one-sided limits.

Left-hand limit (using $x+2$ for $x<1$ ): $\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (x+2) = 1+2 = 3$
Right-hand limit (using $3x$ for $x>1$ ): $\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} 3x = 3(1) = 3$

Since LHL $=$ RHL $=3$ , the limit exists: $\lim_{x \to 1} f(x) = 3$ .

Condition 3 — does the limit equal $f(1)$ ? The limit is $3$ , but $f(1)=5$ . Since $3 \neq 5$ , condition 3 fails. …

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

CBSE 2023Set ANNUAL3 marks
$x = a$
›Reveal solutionSolution
Three conditions: $f(a)$ exists, the limit exists, and the two are equal.

A function $f(x)$ is said to be continuous at $x=a$ if all three of the following conditions are satisfied:

$f(a)$ is defined — the function has a definite, finite value at $x=a$ .

$\lim\limits_{x\to a}f(x)$ exists — the left-hand limit and the right-hand limit are equal and finite:

$\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)$

The limit equals the function value:

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

…
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CBSE 2019Set ANNUAL3 marks
$x = 1$
›Reveal solutionSolution

$f(x)$ is continuous at $x = 1$ since $\lim_{x\to1}f(x) = f(1) = -2$ .

A function is continuous at $x = a$ if (i) $f(a)$ exists, (ii) $\lim_{x\to a}f(x)$ exists, and (iii) the two are equal.

Functional value: given $f(1) = -2$ .

Limit: for $x \neq 1$ , factorise the numerator ( $x^2 - 4x + 3 = (x-1)(x-3)$ ):

$f(x) = \frac{(x-1)(x-3)}{x-1} = x - 3 \quad (x \neq 1).$

Hence

$\lim_{x\to1} f(x) = \lim_{x\to1}(x - 3) = 1 - 3 = -2.$

…

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Q.Check the continuity of the function $f(x) = \begin{cases} x+2, & x<1 \\ 5, & x=1 \\ 3x, & x>1 \end{cases}$ at $x=1$ .

Concept understanding — Continuity of a Function and Its Business Applications

Q.Check the continuity of the function f(x)={x+2,x<15,x=13x,x>1f(x) = \begin{cases} x+2, & x<1 \\ 5, & x=1 \\ 3x, & x>1 \end{cases}f(x)=⎩⎨⎧​x+2,5,3x,​x<1x=1x>1​ at x=1x=1x=1.

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Q.Check the continuity of the function $f(x) = \begin{cases} x+2, & x<1 \\ 5, & x=1 \\ 3x, & x>1 \end{cases}$ at $x=1$ .

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