Imagine drawing the graph of a function and putting your pen down at x=a. If the function is continuous there, you can draw straight through that point without lifting your pen — no jump, no hole, no break. That is the intuition; here is the precision.
The Three-Condition Test
For f(x) to be continuous at x=a, all three must hold. If even one fails, f is discontinuous there.
Important
Continuity at x=a requires:
f(a) is defined,
x→alimf(x) exists (left- and right-hand limits are equal),
x→alimf(x)=f(a).
Condition 1 says a is in the domain — the pen must have somewhere to land. Condition 2 says the curve approaches a single value from both sides — no jump. Condition 3 says that common approach value actually matches the function's value at a — no misplaced point.
Why All Three Are Needed
f(x)=x−1x2−1 has limx→1f(x)=2, yet f(1) is undefined (zero denominator). Condition 1 fails, leaving a hole at (1,2).
A piecewise function shows the opposite can be fine:
f(x)=⎩⎨⎧x+13x+1x<2x=2x>2
Here f(2)=3, both one-sided limits equal 3, and they match f(2) — so all three hold and fis continuous at x=2.
Common Pitfalls
Watch out
"Limit exists" does not mean "continuous." The hole example has a limit but no continuity — the limit must equal the function value.
Watch out
"Defined everywhere" does not mean "continuous." A piecewise function can have a value at every point and still jump. Always check the one-sided limits.
A Quick Check
Is f(x)=∣x∣ continuous at x=0? f(0)=0, limx→0∣x∣=0, and the two agree — yes, even though ∣x∣ has a sharp corner. Continuity demands no break, not smoothness.
Takeaway: continuity at a point means the function value and the two one-sided limits all agree. Agreement ⇒ the graph passes through unbroken; disagreement ⇒ a discontinuity.
Continuity at a Point is the opening idea of the CBSE Class 12 Continuity and Differentiability chapter, and the three-condition test described here matches exactly what NCERT exercises and "continuity and differentiability class 12 important questions" expect students to apply. This concept is also foundational for JEE Main and NEET, where checking continuity is often the first step before testing differentiability of a function.
A function is continuous at a point of its domain when the limit there equals the function value; continuity is only ever discussed at points that actually belong to the domain.
(a)f(x)=x−5 is a polynomial, so it is continuous for every real x.
(b)f(x)=x−51 has domain x=5. On that domain it is a rational function with non-zero denominator, hence continuous at every point of its domain. (x=5 is not in the domain, so continuity is not tested there.)
(c)f(x)=x+5x2−25 has domain x=−5, where it equals x−5; it is continuous at every point of its domain.
(d)f(x)=∣x−5∣ is continuous for every real x.
✓Final answer
All four functions are continuous — each at every point of its domain.
Each function is continuous at every point of its domain: (a) and (d) on all of R, (b) on x=5, and (c) on x=−5.
Continuity is a property we check at points of the domain: f is continuous at x=a (with a in the domain) if limx→af(x)=f(a). A point that is not in the domain is not called a point of discontinuity — the function simply isn't defined there, so there is nothing to test.
(a) f(x)=x−5
A polynomial. For any real a, limx→a(x−5)=a−5=f(a), so f is continuous for all real x.
(b) f(x)=x−51,x=5
The domain is all reals except 5. Take any a=5: the denominator a−5=0, so limx→ax−51=a−51=f(a). Thus f is continuous at every point of its domain. Because 5 is not in the domain, we do not call f "discontinuous at 5".
(c) f(x)=x+5x2−25,x=−5
Factor: x+5x2−25=x+5(x−5)(x+5)=x−5 for x=−5. On its domain f agrees with the polynomial x−5, so for any a=−5, limx→af(x)=a−5=f(a). Hence f is continuous at every point of its domain.
(d) f(x)=∣x−5∣
Absolute value is continuous everywhere. In particular at x=5: limx→5∣x−5∣=0=f(5). The corner at x=5 affects differentiability, not continuity.
✓Final answer
All four functions are continuous — each at every point of its domain: (a) and (d) for all real x, (b) for x=5, (c) for x=−5.
Method: Examining a Function for Continuity When a Restricted Domain Is Given
When a question gives several functions to 'examine for continuity,' the real skill being tested is recognising that continuity is only ever checked at points inside the domain — a function is never called discontinuous at a point where it was never defined in the first place.
Steps
Step 1: Identify the actual domain of each function
Look for any denominator, root, or logarithm that restricts where the function is defined, and note explicitly which real numbers are excluded.
Step 2: For a rational expression, simplify by factoring where valid
If the numerator and denominator share a common factor (e.g. x+ax2−a2=x−a for x=−a), the simplified form describes the function's behaviour everywhere on its domain — but the simplification is only valid where the original denominator is non-zero.
Step 3: Apply the three-condition continuity test at each point of the domain
limx→cf(x)=f(c)for every c in the domain
For a polynomial or a simplified rational/absolute-value expression, direct substitution confirms this at any domain point.
Step 4: State the conclusion in terms of the domain, not 'everywhere'
Report continuity as 'continuous at every point of its domain,' explicitly naming the excluded value(s) rather than saying the function is continuous everywhere or, incorrectly, discontinuous at a point that was never in the domain to begin with.
The general rule: find the domain first, simplify carefully, then apply the continuity test only where the function actually lives.
Common Mistakes
Mistake 1: Calling the function 'discontinuous at x=5' (or x=−5) because the formula isn't defined there
Why it's wrong: a point that is excluded from the domain by the question itself (e.g. x=5) is not a point where continuity is tested at all — 'discontinuous' only applies to domain points where the three-condition test fails, not to points outside the domain. Correct approach: state that the function is continuous at every point of its domain, and note the excluded point separately without calling it a discontinuity.
Mistake 2: Treating x+5x2−25 as identical to x−5 everywhere, including at x=−5
Why it's wrong: the cancellation x+5(x−5)(x+5)=x−5 is only valid where x+5=0; writing the simplified form as if it holds at x=−5 silently changes the function (the original is undefined there, the simplified form isn't). Correct approach: keep the domain restriction x=−5 attached to every statement about the simplified function.