Q.Evaluate: x→1limx−1(x+x2+x3+…+xn)−n
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A Toolkit of Named Limits
Some limits recur so often across problems that it is worth memorising their values outright, along with the one theorem that proves the trickiest of them: the Sandwich (Squeeze) Theorem.
The Sandwich Theorem
Theorem 9.5. If g(x)≤f(x)≤h(x) for all x near x0 (except possibly at x0 itself), and if
limx→x0g(x)=limx→x0h(x)=l,
then limx→x0f(x)=l too — f is "squeezed" between two functions that agree in the limit, so it has no room to do anything else.
Illustration: to show x→0limx2sinx21=0, note that sin(⋅) is always between −1 and 1, so −x2≤x2sinx21≤x2. Since both −x2 and x2 tend to 0 as x→0, the Sandwich Theorem forces the middle expression to 0 as well — even though limx→0sinx21 on its own does not exist (it oscillates wildly), so the product rule alone could never have been applied directly.
This is exactly why the Sandwich Theorem is indispensable rather than a curiosity: whenever one factor oscillates without a limit but is bounded, and the other factor is squeezed to zero, the ordinary product law (Concept 2) is not applicable — you need the sandwich.
The two flagship trigonometric limits
Result 9.1.
(a)limθ→0θsinθ=1(b)limθ→0θ1−cosθ=0
Part (a) is proved geometrically by sandwiching θsinθ between cosθ and 1 using the areas of a triangle, a sector, and a larger triangle built on the unit circle; since both bounding functions tend to 1 as θ→0, so must θsinθ. Part (b) follows algebraically from (a) by writing 1−cosθ=2sin22θ and splitting the quotient into a sin-over-argument piece (which uses part (a)) times a factor that vanishes.
A direct corollary worth keeping separate: x→0limsinx=0, obtained from the sandwich −∣x∣≤sinx≤∣x∣.
The full standard-limit toolkit (§9.2.10)
Alongside the trig pair above, these are worth having on instant recall — none require anything beyond algebra and substitution to use (their proofs, where given, lean on the exponential/log relationship or on the trig pair):
limx→0xex−1=1limx→0xax−1=loga (a>0)limx→0xlog(1+x)=1
limx→0xsin−1x=1limx→0xtan−1x=1
And the three equivalent forms of the number e as a limit:
limx→∞(1+x1)x=elimx→0(1+x)1/x=elimx→∞(1+xk)x=ek
e is a transcendental number — it never satisfies any polynomial equation with rational coefficients. That's part of why it shows up as a genuinely new limiting constant here rather than something expressible in simpler closed form.
The recognise-and-substitute pattern
Nearly every "hard-looking" limit in this section is really one of the above standard forms in disguise, reached via a clean substitution y=(some expression in x) chosen so that y→0 (or y→∞) exactly when x does. The book's worked examples all follow this shape:
- Spot the shell. Identify which standard form the expression resembles — a (1+□)1/□ shape signals e; a □sin(□) shape signals Result 9.1(a); a □a□−1 shape signals Result 9.3.
- Substitute y for the "□" so the expression matches the standard form exactly, tracking what y→ as x→x0.
- Apply the standard limit to the y-expression, then (if the exponent or coefficient outside doesn't vanish) combine using the power/product rules from Concept 2. …
Recognising the expression as the definition of the derivative of f(x)=x+x2+…+xn at x=1 (since f(1)=n), the limit equals $f'(1)=1+2+3+\ldots+n=\df …
The given limit is exactly f′(1) for f(x)=x+x2+…+xn, which works out to 2n(n+1).
Let f(x)=x+x2+x3+…+xn. Then f(1)=1+1+…+1=n (n terms of 1).
The given limit is:
limx→1x−1(x+x2+…+xn)−n=limx→1x−1f(x)−f(1),
which is precisely the definition of the derivative f′(1).
…
Showing the 12 most recent of 31 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.x→0limsin2xsin4x=?(a) 0(b) 2(c) 21(d) 1
›Reveal solutionSolution
Rewrite using the standard limit limθ→0θsinθ=1 applied separately to 4x and 2x.
limx→0sin2xsin4x=limx→02xsin2x×2x4xsin4x×4x=limx→04xsin4x×sin2x2x×2x4x …
- CBSE 2026Set ANNUAL1 markMCQQ.The value of x→0limxtanx is(a) 0(b) 1(c) −1(d) 2
›Reveal solutionSolution
limx→0xtanx=1, using the standard limit limx→0xsinx=1 together with cos0=1.
Write tanx=cosxsinx, so
xtanx=xsinx⋅cosx1
…
- CBSE 2026Set ANNUAL1 markQ.Write value of x→0limxsinx.
›Reveal solutionSolution
x→0limxsinx=1 is a standard limit, provable via the sandwich (squeeze) theorem using the unit circle.
For small x (in radians), geometric comparison of the areas of a triangle, a circular sector, and a larger triangle bounding the unit circle gives sinx<x<tanx for 0<x<2π.
…
- CBSE 2025Set ANNUAL1 markMCQQ.limx→0xsinx=(a) 0(b) 1(c) -1(d) 1/2
›Reveal solutionSolution
limx→0xsinx=1.
This is one of the standard trigonometric limits proved geometrically (comparing areas of triangles and a sector in a unit circle) in the NCERT text. It is used repeatedly to differenti …
- CBSE 2025Set ANNUAL1 markMCQQ.x→0lim(sinbxsinax) is equal to(a) ab(b) ba(c) b−a(d) a−b
›Reveal solutionSolution
As x→0, sin(ax)≈ax and sin(bx)≈bx, so the ratio tends to a/b.
limx→0sinbxsinax=limx→0bxsinbx⋅bxaxsinax⋅ax=limx→0bxsinbxaxsinax⋅ba
…
- CBSE 2025Set ANNUAL1 markMCQQ.x→0lim(23x−132x−1) is equal to(a) log8log9(b) log9log8(c) log3log2(d) log2log3
›Reveal solutionSolution
Rewriting 32x=9x and 23x=8x and applying the standard limit limx→0xkx−1=logk to numerator and denominator gives log8log9.
limx→023x−132x−1=limx→08x−19x−1
Divide numerator and denominator by x:
=limx→0x8x−1x9x−1
…
- CBSE 2025Set ANNUAL1 markQ.Write the value of x→0limxtanx.
›Reveal solutionSolution
The standard limit x→0limxtanx=1 follows from x→0limxsinx=1 and x→0limcosx=1.
xtanx=xsinx⋅cosx1.
…
- CBSE 2024Set ANNUAL1 markMCQQ.x→πlimπ(π−x)sin(π−x)=(a) π1(b) π21(c) 1(d) None of these
›Reveal solutionSolution
Substitute t=π−x to turn this into the standard limit limt→0tsint=1.
Let t=π−x. As x→π, t→0. Rewrite the limit:
limx→ππ(π−x)sin(π−x)=limt→0πtsint=π1limt→0tsint
…
- CBSE 2024Set ANNUAL1 markQ.Find: lim(x→0) (a^x - 1) / x = ?
›Reveal solutionSolution
This is a standard limit result: limx→0xax−1=lna (natural log of a).
Write ax=exlna. Then
xax−1=xexlna−1=lna⋅xlnaexlna−1.
…
- CBSE 2024Set ANNUAL1 markMCQQ.x→0limxax−bx=(a) log(ab)(b) logab(c) ba(d) log(ba)
›Reveal solutionSolution
The limit equals log(ba).
Write the limit as a difference of two standard limits:
limx→0xax−bx=limx→0xax−1−limx→0xbx−1=loga−logb=log(ba), …
- CBSE 2024Set ANNUAL1 markMCQQ.The value of x→0limxex−1=……(a) 0(b) 1(c) ∞(d) ex.
›Reveal solutionSolution
x→0limxex−1=1.
Using the series expansion ex=1+x+2!x2+3!x3+⋯, we get
ex−1=x+2!x2+3!x3+⋯,
so
xex−1=1+2!x+3!x2+⋯. …
- CBSE 2024Set ANNUAL1 markMCQQ.The value of x→0limxtanx=……(a) x(b) 1(c) 0(d) ∞.
›Reveal solutionSolution
x→0limxtanx=1.
Write tanx=cosxsinx, so
xtanx=xsinx⋅cosx1. …
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