Q.Prove that the function f(x)=xn is continuous at x=n, where n is a positive integer.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Continuity At A Point
Continuity at a Point
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.
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 f is continuous at x=2.
Common Pitfalls
"Limit exists" does not mean "continuous." The hole example has a limit but no continuity — the limit must equal the function value.
"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 …
Concept: Continuity At A Point — f(x)=xn is continuous at x=n if limx→nxn=nn.
Step 1: Factor the difference of nth powers:
xn−nn=(x−n)(xn−1+xn−2n+⋯+xnn−2+nn−1).
Step 2: Call the second factor S(x)=xn−1+xn−2n+⋯+nn−1. It is a sum of products of the continuous identity function x with constants, hence S(x) is continuous everywhere, so
limx→nS(x)=S(n)=n termsnn−1+nn−1+⋯+nn−1=n⋅nn−1=nn.
Step 3: So xn=nn+(x−n)S(x), and taking the limit as x→n: …
Using the factorisation xn−nn=(x−n)(xn−1+xn−2n+⋯+nn−1), the second factor is a continuous (polynomial-type) expression whose limit at x=n is nn; since the (x−n) factor goes to 0, limx→nxn=nn=f(n), so f(x)=xn is continuous at x=n. No epsilon-delta is needed — this is the chapter's own limit/algebraic method.
Setting Up the Proof
To prove f(x)=xn is continuous at x=n (for a positive integer n), we must show all three continuity conditions hold:
- f(n)=nn is defined.
- limx→nf(x) exists.
- limx→nf(x)=f(n).
The first is immediate. The real work is showing limx→nxn=nn, and the cleanest in-syllabus way to do that is an algebraic factorisation, not an epsilon-delta chase.
Step-by-Step Reasoning
1. Factor xn−nn using the standard difference-of-powers identity.
For any real x and positive integer n:
xn−nn=(x−n)(xn−1+xn−2n+xn−3n2+⋯+xnn−2+nn−1).
This identity is exact and holds for every x, not just x near n — it comes from expanding the right-hand side and watching every middle term cancel in a telescoping pattern.
2. Name the second factor S(x).
Let
S(x)=xn−1+xn−2n+xn−3n2+⋯+xnn−2+nn−1.
Each term of S(x) is a constant times a power of x, so S(x) is built entirely from the continuous identity function x using multiplication by constants and addition — by the algebra of continuous functions, S(x) is continuous for every real x, in particular at x=n.
3. Evaluate S(n).
Substituting x=n into every term gives n identical terms, each equal to nn−1:
S(n)=nn−1+nn−1+⋯+nn−1 (n terms)=n⋅nn−1=nn.
Since S is continuous at n, limx→nS(x)=S(n)=nn.
4. Rewrite xn using the factorisation and take the limit.
From Step 1, for every x:
xn=nn+(x−n)S(x).
Taking x→n on both sides, and using that a product of limits is the limit of the product (algebra of limits):
limx→nxn=nn+(limx→n(x−n))(limx→nS(x))=nn+(0)(nn)=nn.
5. Conclude. …
Method: Proving f(x)=xn Is Continuous at x=n Using Limit Laws
The standard Class 12 way to prove continuity of f(x)=xn at a point is an algebraic limit argument built on the factorisation of xn−nn — not a formal epsilon-delta argument, which is outside the NCERT Class 12 syllabus for this chapter.
Steps
Step 1: State what must be shown
To prove continuity at x=n, it suffices to show
limx→nf(x)=f(n)=nn.
Step 2: Factor the difference xn−nn
Use the standard algebraic identity
xn−nn=(x−n)(xn−1+xn−2n+⋯+xnn−2+nn−1),
valid for any positive integer n.
Step 3: Take the limit using the algebra-of-limits rules
Since each term xn−1−knk is itself a polynomial in x, the sum xn−1+xn−2n+⋯+nn−1 is continuous, so as x→n it tends to n⋅nn−1=nn; and (x−n)→0. By the product rule for limits, …
Common Mistakes
Mistake 1: Trying to prove continuity using a formal epsilon-delta argument
Why it's wrong: NCERT Class 12 Continuity and Differentiability builds continuity proofs on the algebra of limits (limits of sums, products, and known limits of polynomials), not on a formal epsilon-delta construction — that is a university-level technique outside this syllabus and unnecessary complexity for what the algebraic factorisation already proves cleanly. Correct approach: factor xn−nn=(x−n)(xn−1+xn−2n+⋯+nn−1) and take the limit using the standard limit laws for sums and products.
Mistake 2: Misremembering the factorisation of xn−nn …
Showing the 12 most recent of 32 on this concept.
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.If f(x)=⎩⎨⎧(2+5x22+3x2)x28+3,k,for x=0for x=0 is a continuous function at x=0, then k= (A) e−2 (B) e−4 (C) e−8 (D) e−16
›Reveal solutionSolution
This tests the standard 1∞ exponential limit technique. Taking logs and expanding to first order in x2 gives k=e−8.
Concept and Intuition
Expressions of the form (base →1)power→∞ are evaluated by writing L=elim(power)⋅log(base) and expanding log(1+u)≈u for the small quantity u= base−1. The constant added to the exponent (here +3) never survives, since it multiplies a term that itself →0.
Step-by-Step Solution
- For continuity at x=0: k=limx→0(2+5x22+3x2)8/x2+3.
- Let L denote this limit. Then logL=limx→0(x28+3)log(2+5x22+3x2).
- Write the base as 1+u where u=2+5x22+3x2−1=2+5x2−2x2≈−x2 for small x (to leading order).
- log(1+u)≈u≈−x2 for small x. …
- AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQQ.If the function f(x)=x1+x−1 is continuous at x=0 then f(0)= (A) −21 (B) 31 (C) 21 (D) −31
›Reveal solutionSolution
Rationalizing the numerator turns the 0/0 form into a simple limit, which must equal f(0) for continuity.
Concept and Intuition
For f to be continuous at x=0, we need f(0)=x→0limf(x). Since the given formula for f(x) is 0/0 at x=0, we must simplify it algebraically first.
Step-by-Step Solution
- x1+x−1=x(1+x+1)(1+x−1)(1+x+1)=x(1+x+1)(1+x)−1=x(1+x+1)x=1+x+11.
- As x→0: 1+11=21. …
- AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQQ.If a function defined by f(x)=sinxlog(1+x)(3x−1)2, x=0, is continuous at x=0, then f(0)= (A) 2log3 (B) log32 (C) 2+log3 (D) (log3)2
›Reveal solutionSolution
Standard small-x equivalents give f(x)→(ln3)2 as x→0, so f(0)=(log3)2.
Concept and Intuition
For continuity at x=0, f(0) must equal limx→0f(x). Use the standard limits limx→0xax−1=lna, limx→0xsinx=1, limx→0xlog(1+x)=1.
Step-by-Step Solution
- (3x−1)2=x2(x3x−1)2→x2(ln3)2 as x→0.
- sinxlog(1+x)=x⋅xsinx⋅x⋅xlog(1+x)=x2⋅xsinx⋅xlog(1+x)→x2 as x→0.
- So f(x)=sinxlog(1+x)(3x−1)2→x2x2(ln3)2=(ln3)2. …
- AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQQ.If f:R→R defined by f(x)=⎩⎨⎧xsinx−sin2x,x3/2x2+x−x,x<0x>0 is continuous on R, then f(0)= (A) 1/2 (B) 3/2 (C) 1 (D) −1
›Reveal solutionSolution
This tests continuity of a piecewise function at the junction point using standard small-x approximations; both one-sided limits equal 1/2.
Concept and Intuition
A piecewise function is continuous at a boundary point only if the left-hand limit, right-hand limit, and the defined value at that point all agree. Here f(0) isn't given directly by either branch (both blow up as 0/0), so we must compute the limits from each side and set f(0) equal to their common value.
Step-by-Step Solution
- Left-hand limit (x→0−):
limx→0−xsinx−sin(x/2)
Use the small-angle expansion sint=t−6t3+⋯:
sinx−sin2x=(x−6x3)−(2x−48x3)+⋯=2x+O(x3)
Dividing by x: the limit is 21.
- Right-hand limit (x→0+):
limx→0+x3/2x2+x−x=limx→0+x3/2x(x+1−1)=limx→0+xx+1−1
Using 1+x≈1+2x−8x2+⋯:
xx+1−1≈xx/2=21 …
- AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQQ.If f(x)=sinx3loge(1+x2(tanx)), x=0 is to be continuous at x=0, then f(0) must be equal to ______ (A) 1 (B) 0 (C) 21 (D) −1
›Reveal solutionSolution
Using the small-angle equivalences loge(1+u)∼u, tanx∼x, and sinx3∼x3 near x=0, the limit of f(x) works out to 1, so f(0)=1 is required for continuity.
Concept and Intuition
For a function to be continuous at a removable-looking point like x=0, its value there must be defined equal to the limit of the function as x→0. Standard small-x equivalences (loge(1+u)∼u for small u; tanx∼x; sinx∼x) let us evaluate the 00-type limit cleanly.
Step-by-Step Solution
- As x→0, let u=x2tanx. Since tanx∼x for small x, u∼x2⋅x=x3→0.
- Using loge(1+u)∼u for small u: loge(1+x2tanx)∼x2tanx∼x3.
- For the denominator, sin(x3)∼x3 as x→0 (since sinθ∼θ for small θ, here θ=x3→0).
- So f(x)=sin(x3)loge(1+x2tanx)→x3x3=1 as x→0. …
- AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQQ.If the function f(x), defined below, is continuous everywhere, then 'k' equals ______ f(x)=⎩⎨⎧1+x−12x−1,k,−1≤x<∞,x=0x=0 (A) 21loge2 (B) loge4 (C) loge8 (D) loge2
›Reveal solutionSolution
Continuity at x=0 requires k to equal the limit of f(x) as x→0, which works out to ln4 using standard small-x approximations.
Concept and Intuition
For f to be continuous at x=0, we need x→0limf(x)=f(0)=k. Both numerator and denominator vanish as x→0 (this is a 0/0 form), so we use the standard first-order approximations 2x−1≈xln2 and 1+x−1≈x/2 near x=0.
Step-by-Step Solution
- Numerator: 2x−1=exln2−1≈xln2 for small x.
- Denominator: 1+x−1=(1+x)1/2−1≈2x for small x (binomial expansion).
- So x→0lim1+x−12x−1=x→0limx/2xln2=2ln2. …
- AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQQ.Let Sn=1+3x+9x2+27x3+…n terms and −31<x<31. If limn→∞Sn=f(x), then f(x) is discontinuous at the point x= (A) 0 (B) 31 (C) 1 (D) −1
›Reveal solutionSolution
The infinite geometric series sums to f(x)=1/(1−3x) on (−1/3,1/3), and this function has an infinite discontinuity exactly at x=1/3, the edge of the interval where the series stops converging.
Concept and Intuition
An infinite geometric series 1+r+r2+… converges to 1−r1 only when ∣r∣<1; as r→1 this sum diverges to infinity. Here r=3x, so the series converges for ∣x∣<1/3 and the resulting closed-form function has a vertical asymptote right at the boundary x=1/3.
Step-by-Step Solution
- Sn=1+3x+(3x)2+⋯+(3x)n−1, a geometric series with first term 1 and common ratio 3x.
- Sum: Sn=1−3x1−(3x)n (for 3x=1).
- For −31<x<31, ∣3x∣<1, so (3x)n→0 as n→∞.
- Hence f(x)=limn→∞Sn=1−3x1.
- This function f(x)=1−3x1 is undefined/blows up exactly where 1−3x=0, i.e. x=31 — an infinite discontinuity right at the edge of the domain of convergence. …
- AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQQ.If f(x)=x2cos4x−1, x=0 and f(0)=2 is a real valued function, then (A) x→0limf(x) does not exist (B) x→0limf(x)=1 (C) f is not continuous at x = 0 (D) f is continuous at x = 0
›Reveal solutionSolution
This tests factoring cos4x−1 and using the standard limit xsinx→1; the limit exists and equals −2, but f(0)=2, so f fails continuity at 0.
Concept and Intuition
cos4x−1 factors as a difference of squares twice: cos4x−1=(cos2x−1)(cos2x+1)=−sin2x(cos2x+1). This lets us isolate the familiar (xsinx)2→1 building block.
Step-by-Step Solution
- cos4x−1=(cos2x−1)(cos2x+1)=−sin2x(cos2x+1).
- f(x)=x2−sin2x(cos2x+1)=−(xsinx)2(cos2x+1).
- As x→0: (xsinx)2→1 and cos2x+1→2.
- So x→0limf(x)=−1×2=−2 — this limit exists (rules out (A)) and is not 1 (rules out (B)). …
- AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQQ.If the function f defined by f(x)=⎩⎨⎧x21−cos4x,a,16+x−4x,x<0x=0x>0 is continuous at x=0, then a= (A) 1 (B) 2 (C) 4 (D) 8
›Reveal solutionSolution
Continuity at x=0 requires the left limit, the right limit, and f(0)=a to all agree; both one-sided limits work out to 8. Answer: a=8.
Concept and Intuition
For a piecewise function to be continuous at a junction point, the value defined there (a) must equal both the limit approaching from the left and the limit approaching from the right. Each side here needs a standard trick: the trig side uses the double-angle identity 1−cosθ=2sin2(θ/2), and the surd side needs rationalisation to remove the − indeterminate form.
Step-by-Step Solution
- Left-hand limit (x→0−): 1−cos4x=2sin2(2x), so
limx→0−x21−cos4x=limx→0−x22sin2(2x)=2limx→0(2xsin2x)2⋅4=2⋅1⋅4=8.
- Right-hand limit (x→0+): rationalise 16+x−4x by multiplying top and bottom by 16+x+4: (16+x)−16x(16+x+4)=xx(16+x+4)=16+x+4. …
- AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQQ.If a real valued function f(x)=⎩⎨⎧(1+sinx)cosecx,a,ae2/x+be3/xe2/x+e3/x,−π/2<x<0x=00<x<π/2 is continuous at x = 0, then ab= (A) e (B) e2 (C) 1 (D) −1
›Reveal solutionSolution
The left-hand limit is the classical 1∞ form giving e, fixing a=e;
the right-hand limit needs dividing by the dominant exponential to fix b.
Continuity forces both, and ab=1.
Concept and Intuition
A piecewise function is continuous at a point only if the left-hand limit,
right-hand limit, and the function's value there all agree. Here the left
piece is a 1∞ indeterminate form (standard trick: exponentiate and use
log(1+u)∼u), and the right piece is a ratio of two exponentials growing at
different rates as x→0+ (since 1/x→+∞), so the faster-growing
exponential e3/x dominates and everything else becomes negligible after
dividing through by it.
Step-by-Step Solution
- Left limit: L=limx→0−(1+sinx)cosecx. Take logs: logL=limcosecx⋅log(1+sinx)=limsinxsinx(1+O(sinx))→1. So L=e. Continuity requires f(0)=a=L=e.
- Right limit: R=limx→0+ae2/x+be3/xe2/x+e3/x. Divide numerator and denominator by e3/x: R=limx→0+ae−1/x+be−1/x+1. …
- AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQQ.If f(x)=⎩⎨⎧sin2x(eax−1)log(1+x),2,tan2xcos4x−cosbx,if x>0if x=0if x<0 is continuous at x=0 then b2−a2= (A) 4 (B) 5 (C) 3 (D) 7
›Reveal solutionSolution
Continuity at x=0 forces both one-sided limits to equal f(0)=2; solving gives a=2, b2=20, so b2−a2=4.
Concept and Intuition
For f to be continuous at 0, we need x→0+limf(x)=x→0−limf(x)=f(0)=2. Each one-sided piece is a 0/0 form that resolves using the standard small-angle equivalences et−1∼t, log(1+t)∼t, sint∼t, tant∼t, and cost≈1−2t2.
Step-by-Step Solution
- Right-hand limit (x→0+): using eax−1∼ax, log(1+x)∼x, sin2x∼x2,
limx→0+sin2x(eax−1)log(1+x)=limx→0+x2(ax)(x)=a.
Setting this equal to f(0)=2: a=2.
2. Left-hand limit (x→0−): expand cos4x≈1−2(4x)2=1−8x2 and cos(bx)≈1−2b2x2, and tan2x∼x2:
cos4x−cos(bx)≈(1−8x2)−(1−2b2x2)=x2(2b2−8). …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.If a function f defined by f(x)=⎩⎨⎧x21−cos4x,a,16+x−4x,x<0x=0x>0 is continuous at x=0, then a= (A) 8 (B) 4 (C) 3 (D) 2
›Reveal solutionSolution
Both one-sided limits at x=0 evaluate to 8 (via 1−cosθ=2sin2(θ/2) on the left, and rationalizing on the right), so a=8.
Concept and Intuition
For f to be continuous at 0, the left-hand limit, right-hand limit, and f(0)=a must all agree. The left piece is a classic 0/0 trig limit solved via the half-angle identity for 1−cosθ; the right piece is a classic surd limit solved by rationalizing the denominator.
Step-by-Step Solution
- Left limit: 1−cos4x=2sin2(2x), so x21−cos4x=x22sin2(2x)=8(2xsin2x)2.
- As x→0−, 2xsin2x→1, so this limit →8(1)2=8.
- Right limit: let t=x (so t→0+ as x→0+): 16+x−4x=16+t−4t.
- Rationalize: multiply by 16+t+416+t+4: (16+t)−16t(16+t+4)=tt(16+t+4)=16+t+4. …
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