Chapter 14: Limits, Continuity and Differentiability (Set-4)
Evaluate limx→0sin2xsin5xlimx→0sin5xsin2x
A 5/2
B 10
C 0
D 2/5
For small x, sinkx∼kxsinkx∼kx. So sin2xsin5x∼2x5x=2/5sin5xsin2x∼5x2x=2/5. This uses standard small-angle limit behavior.
Evaluate limx→0ex−e−x2xlimx→02xex−e−x
A 0
B 1
C 2
D e
Use expansions: ex=1+x+…ex=1+x+…, e−x=1−x+…e−x=1−x+…. Difference is 2x+…2x+…. Dividing by 2x gives 1.
Evaluate limx→0ln(1+x)−xx2limx→0x2ln(1+x)−x
A 1/2
B 0
C −1/2
D −1
Series: ln(1+x)=x−x22+…ln(1+x)=x−2×2+…. So ln(1+x)−x≈−x22ln(1+x)−x≈−2×2. Dividing by x2x2 gives −1/2.
Evaluate limx→0ax−1xlimx→0xax−1 for a=10
A ln 10
B 10
C 1/10
D 0
Use ax=exlnaax=exlna. Then ax−1x=exlna−1x→lnaxax−1=xexlna−1→lna. For a=10, the limit is ln 10.
Evaluate limx→01+x−1−xxlimx→0x1+x−1−x
A 0
B −1
C 1
D 2
Rationalize: 1+x−1−xx⋅1+x+1−x1+x+1−x=21+x+1−xx1+x−1−x⋅1+x+1−x1+x+1−x=1+x+1−x2. As x→0, denominator→2, so limit 1.
Evaluate limx→0cosx−cos3xx2limx→0x2cosx−cos3x
A −4
B 2
C 0
D 4
Use cosu≈1−u22cosu≈1−2u2. Then cosx−cos3x≈(1−x22)−(1−9×22)=4x2cosx−cos3x≈(1−2×2)−(1−29×2)=4×2. Divide by x2x2 gives 4.
Evaluate limx→0sinx−xcosxx3limx→0x3sinx−xcosx
A 1/2
B 1/3
C 0
D −1/3
Use expansions: sinx=x−x36sinx=x−6×3, cosx=1−x22cosx=1−2×2. Then xcosx=x−x32xcosx=x−2×3. Difference sinx−xcosx≈(−x36+x32)=x33sinx−xcosx≈(−6×3+2×3)=3×3. Ratio→1/3.
Evaluate limx→0tanx−xx3limx→0x3tanx−x
A 1/2
B 1/6
C 1/3
D 0
Series: tanx=x+x33+…tanx=x+3×3+…. So tanx−x≈x33tanx−x≈3×3. Dividing by x3x3 gives 1/3.
Evaluate limx→01x(11+x−1)limx→0x1(1+x1−1)
A 1
B 0
C −2
D −1
Simplify: 1x(1−(1+x)1+x)=1x(−x1+x)=−11+xx1(1+x1−(1+x))=x1(1+x−x)=1+x−1. As x→0, limit is −1.
Evaluate limx→0(1+x)5−1xlimx→0x(1+x)5−1
A 1
B 5
C 0
D 10
Use binomial expansion: (1+x)5=1+5x+10×2+…(1+x)5=1+5x+10×2+…. Subtract 1 and divide by x gives 5+10x+…5+10x+…. As x→0, limit is 5.
Evaluate limx→0(1+x)α−1xlimx→0x(1+x)α−1
A 1
B 0
C αα
D α2α2
Using expansion (1+x)α=1+αx+O(x2)(1+x)α=1+αx+O(x2). Subtract 1 and divide by x gives α+O(x)α+O(x). Hence the limit equals αα.
Evaluate limx→0sinx⋅sin2xx2limx→0x2sinx⋅sin2x
A 1
B 2
C 0
D 4
Write sinxx⋅sin2xx=(sinxx)(sin2x2x)⋅2xsinx⋅xsin2x=(xsinx)(2xsin2x)⋅2. Both bracketed limits are 1, so result is 2.
Use standard limits: sinxx→1xsinx→1 and 1−cosxx2→1/2×21−cosx→1/2. Product tends to 1⋅1/2=1/21⋅1/2=1/2.
If f is continuous at a and f(a)≠0, then 1ff1 is continuous at a because
A product property
B jump property
C reciprocal property
D squeeze property
If f is continuous at a and not zero there, values near a stay away from 0. Hence 1/f is defined near a and remains continuous by continuity algebra.
If f and g are continuous at a, then max(f,g)max(f,g) is
A continuous at a
B discontinuous always
C differentiable always
D undefined at a
max(f,g) can be expressed using max(f,g)=f+g+∣f−g∣2max(f,g)=2f+g+∣f−g∣. If f and g are continuous, so are f−g and |f−g|, hence max is continuous.
A function can be continuous at a but not differentiable at a due to
A removable hole
B vertical asymptote
C jump break
D corner point
At a corner, the function joins smoothly without a break, so continuity can hold. But left and right slopes differ, so derivative does not exist.
For f(x)=∣x∣f(x)=∣x∣, second derivative at x=0 is
A exists finite
B does not exist
C equals 0
D equals 1
|x| is not differentiable at 0, so its first derivative is not defined there. Without a first derivative at 0, the second derivative at 0 cannot exist.
If a function has oscillatory discontinuity at 0, then usually
A LHL exists
B RHL exists
C limit fails
D value equals limit
Oscillation near a point prevents approaching a single number. Even if the function is bounded, rapid oscillations can stop the limit from existing.
If limx→af(x)=Llimx→af(x)=L and limx→ag(x)=Mlimx→ag(x)=M, then lim(f−g)lim(f−g) equals
A L−M
B LM
C L/M
D M−L
By difference law, lim(f−g)=limf−limglim(f−g)=limf−limg, provided both limits exist. This simplifies many algebraic limit evaluations.
Evaluate limx→0sin4xtan2xlimx→0tan2xsin4x
A 1
B 1/2
C 4
D 2
sin4xtan2x=sin4x4x⋅2xtan2x⋅2tan2xsin4x=4xsin4x⋅tan2x2x⋅2. Each limit tends to 1, so total tends to 2.
Evaluate limx→0tan3xsin2xlimx→0sin2xtan3x
A 2/3
B 6
C 3/2
D 1
For small x, tan3x∼3xtan3x∼3x and sin2x∼2xsin2x∼2x. Ratio tends to 3x2x=3/22x3x=3/2.
Use 1−cosu∼u221−cosu∼2u2. Then numerator ∼(2x)22=2×2∼2(2x)2=2×2, denominator ∼x22∼2×2. Ratio ∼2x2x2/2=4∼x2/22×2=4.
Evaluate limx→0sin23xsin2xlimx→0sin2xsin23x
A 3
B 1/9
C 0
D 9
sin3xsinx→3sinxsin3x→3 as x→0 using small-angle behavior. Squaring gives (sin3xsinx)2→9(sinxsin3x)2→9.
Evaluate limx→0sinxx⋅tanxxlimx→0xsinx⋅xtanx
A 0
B 2
C 1
D 1/2
sinxx→1xsinx→1. Also tanxx=sinxx⋅1cosx→1⋅1=1xtanx=xsinx⋅cosx1→1⋅1=1. Product tends to 1.
Evaluate limx→0sinxx⋅1cosxlimx→0xsinx⋅cosx1
A 1
B 0
C 2
D ∞
As x→0, sinxx→1xsinx→1 and cosx→1cosx→1. Hence 1cosx→1cosx1→1. Product tends to 1.
If f is differentiable at a, then limh→0f(a+h)−f(a)hlimh→0hf(a+h)−f(a) equals
A f(a)
B f′(a)
C 0
D ∞
This limit is exactly the definition of derivative at a using first principles. If the derivative exists, this difference quotient limit equals f′(a).
If left derivative equals right derivative at a, then function is
A discontinuous at a
B oscillatory at a
C infinite at a
D differentiable at a
Equality of left and right derivatives implies a single slope exists from both sides, so the derivative exists at that point. Continuity is also required and follows if derivative exists.
For y=x2cosxy=x2cosx, y′ uses
A quotient rule
B only power rule
C product rule
D only chain rule
The function is product of x² and cos x. Differentiate as 2xcosx+x2(−sinx)2xcosx+x2(−sinx). Product rule ensures both parts are differentiated correctly.
For y=sinxxy=xsinx, derivative needs
A only chain rule
B quotient rule
C only power
D no rule
sinxsinx divided by x requires quotient rule: xcosx−sinxx2x2xcosx−sinx. This gives correct derivative for x≠0.
Evaluate limx→0sinx−tanxx3limx→0x3sinx−tanx
A 1/2
B −1/3
C 0
D −1/2
Series: sinx=x−x36sinx=x−6×3, tanx=x+x33tanx=x+3×3. Difference sinx−tanx≈−x36−x33=−x32sinx−tanx≈−6×3−3×3=−2×3. Divide by x3x3 gives −1/2.
Evaluate limx→0ln(1+2x)−2xx2limx→0x2ln(1+2x)−2x
A −1
B −1/2
C −2
D 0
ln(1+u)=u−u22+…ln(1+u)=u−2u2+…. With u=2x, ln(1+2x)=2x−(2x)22+…ln(1+2x)=2x−2(2x)2+…=2x−2×2+…2x−2×2+…. Subtract 2x gives −2x², divide by x² → −2.
Evaluate limx→0ex−1−xx2limx→0x2ex−1−x
A 1/2
B 1
C 0
D 2
Expansion: ex=1+x+x22+…ex=1+x+2×2+…. Then ex−1−x≈x22ex−1−x≈2×2. Dividing by x2x2 gives 1/2.
Discontinuity of tanxtanx at x=π/2x=π/2 is
A removable type
B jump type
C oscillatory type
D infinite type
tanx=sinxcosxtanx=cosxsinx blows up as cosx→0cosx→0 near π/2π/2. Values become unbounded, indicating an infinite discontinuity.
A function continuous on [a,b] must be
A always linear
B bounded
C always constant
D always periodic
A standard property is that a function continuous on a closed interval [a,b] attains maximum and minimum values, hence it is bounded on that interval.
A continuous function on [a,b] always attains
A only maximum
B only minimum
C maximum and minimum
D neither
By the extreme value theorem, continuity on a closed interval guarantees the function reaches both its greatest and least values somewhere in [a,b].
If f is continuous on [a,b] and f(a)f(b) < 0, then
A no root
B jump occurs
C derivative zero
D root exists
By Intermediate Value Theorem, if f changes sign between a and b and is continuous, then there is some c in (a,b) with f(c)=0.
Rolle’s theorem conclusion gives
A f is constant
B some c with f′(c)=0
C f has jump
D limit diverges
If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), Rolle’s theorem guarantees at least one point c where the tangent is horizontal.
For x∣x∣∣x∣x at x=0, which limit exists
A only left
B only right
C neither side
D both equal
As x→0⁺, x/|x|=1; as x→0⁻, x/|x|=−1. One-sided limits exist but are unequal, so two-sided limit does not exist at 0.
For f(x)={x2,x≤12x−1,x>1f(x)={x2,2x−1,x≤1x>1, continuity at 1 needs
A 1=1
B derivative equal
C slopes infinite
D LHL=RHL
At x=1, LHL from x² is 1, RHL from 2x−1 is also 1, and f(1)=1. So LHL=RHL=f(1) ensures continuity.
For the same piecewise function, differentiability at 1 needs
A values equal
B slopes equal
C jump exists
D hole exists
Differentiate each side: for x≤1, derivative is 2x so at 1 gives 2; for x>1, derivative is 2. Left and right derivatives match, so differentiable at 1.
Using Leibnitz theorem, (fg)(2)(fg)(2) equals
A f″g +2f′g′+fg″
B f″g″
C f′g′ only
D f″g + fg″
Leibnitz formula for n=2 gives (fg)′′=(20)fg′′+(21)f′g′+(22)f′′g(fg)′′=(02)fg′′+(12)f′g′+(22)f′′g, which simplifies to fg′′+2f′g′+f′′gfg′′+2f′g′+f′′g.
If f(x)=x3f(x)=x3 and g(x)=exg(x)=ex, then in Leibnitz, g(n)g(n) is
A x^n
B n e^x
C 0
D e^x
All derivatives of e^x are e^x. So g(n)(x)=exg(n)(x)=ex for every n. This makes Leibnitz expansions easier for products with e^x.
Evaluate limx→0(1+x)1/xlimx→0(1+x)1/x equals
A 1
B e
C 0
D ∞
A classic limit: limx→0(1+x)1/x=elimx→0(1+x)1/x=e. It comes from the definition of e or by taking logs and using ln(1+x)∼xln(1+x)∼x.
Evaluate limx→∞(1+1x)xlimx→∞(1+x1)x equals
A 1
B 0
C e
D ∞
Another standard definition of e is limx→∞(1+1x)x=elimx→∞(1+x1)x=e. It is a 1^∞ type indeterminate form.
For indeterminate ∞−∞, a common fix is
A separate limits
B replace by ∞
C replace by 0
D rationalize
Expressions like x2+1−xx2+1−x give ∞−∞. Multiplying by conjugate converts it into a quotient that can be simplified and evaluated.
Evaluate limx→∞(x2+1−x)limx→∞(x2+1−x)
A 1/2
B 1
C 0
D ∞
Rationalize to get x2+1−x=1×2+1+xx2+1−x=x2+1+x1. As x→∞, the denominator grows like 2x, so the expression tends to 0.
Evaluate limx→01+x−1−x2x2limx→0x21+x−1−2x
A 1/8
B −1/8
C 0
D −1/4
Use expansion 1+x=1+x2−x28+…1+x=1+2x−8×2+…. Subtract 1+x21+2x leaves −x28−8×2. Divide by x2x2 gives −1/8.
If f−1f−1 exists and f is differentiable at a with f′(a)≠0, then (f−1)′(f(a))(f−1)′(f(a)) equals
A 1/f′(a)
B f′(a)
C f(a)
D 0
The derivative of inverse function satisfies (f−1)′(y)=1f′(x)(f−1)′(y)=f′(x)1 where y=f(x). So at y=f(a), derivative equals 1/f′(a), provided f′(a)≠0.
If f is increasing and differentiable, then f′(x) is
A ≤ 0
B always 0
C undefined
D ≥ 0
A differentiable increasing function has nonnegative derivative over the interval. Positive derivative means strictly increasing locally; zero derivative can occur at flat points while still increasing overall.
A basic condition for L’Hospital use is
A 0·∞ only
B 0/0 or ∞/∞
C always applicable
D only polynomials
L’Hospital’s rule applies when a limit produces 0/0 or ∞/∞ and numerator/denominator are differentiable near the point. After differentiating, we re-check the new limit.