Chapter 14: Limits, Continuity and Differentiability (Set-3)
Evaluate limx→2×2−4x−2limx→2x−2×2−4
A 0
B 4
C 2
D 8
Factor x2−4=(x−2)(x+2)x2−4=(x−2)(x+2). Cancel (x−2) for x≠2, so limit becomes limx→2(x+2)=4limx→2(x+2)=4. This is a removable-type simplification.
Evaluate limx→0sin3xxlimx→0xsin3x
A 3
B 1
C 0
D 1/3
Write sin3xx=sin3x3x⋅3xsin3x=3xsin3x⋅3. As x→0, sin3x3x→13xsin3x→1. Hence limit is 3.
Evaluate limx→0tan5xtanxlimx→0tanxtan5x
A 1/5
B 0
C 5
D 1
Use tankx∼kxtankx∼kx for small x. Then tan5xtanx∼5xx=5tanxtan5x∼x5x=5. Standard small-angle behavior gives the result.
Evaluate limx→01−cos4xx2limx→0x21−cos4x
A 4
B 16
C 0
D 8
Use 1−cosu∼u221−cosu∼2u2. With u=4x, numerator ∼(4x)22=8×2∼2(4x)2=8×2. Divide by x2x2 to get 8.
Evaluate limx→0e2x−1xlimx→0xe2x−1
A 1
B e
C 2
D 0
Use standard limit limx→0eax−1x=alimx→0xeax−1=a. Here a=2, so the limit is 2. Series e2x=1+2x+…e2x=1+2x+… also confirms.
Evaluate limx→0ln(1+3x)xlimx→0xln(1+3x)
A 3
B 1
C 0
D 1/3
Use ln(1+u)∼uln(1+u)∼u near 0. Here u=3x, so ln(1+3x)x∼3xx=3xln(1+3x)∼x3x=3. Hence limit equals 3.
Evaluate limx→01+x−1xlimx→0x1+x−1
A 1
B 2
C 1/2
D 0
Rationalize: 1+x−1x⋅1+x+11+x+1=11+x+1×1+x−1⋅1+x+11+x+1=1+x+11. As x→0, denominator→2, so limit 1/2.
Evaluate limx→01+2x−1xlimx→0x1+2x−1
A 1
B 2
C 1/2
D 0
Rationalize: 1+2x−1x=21+2x+1×1+2x−1=1+2x+12. As x→0, denominator→2, giving limit 2/2=12/2=1.
Evaluate limx→1×3−1x−1limx→1x−1×3−1
A 1
B 3
C 2
D 0
Factor x3−1=(x−1)(x2+x+1)x3−1=(x−1)(x2+x+1). Cancel (x−1), then substitute x=1: 1+1+1=31+1+1=3. So the limit equals 3.
Evaluate limx→0sinxtanxlimx→0tanxsinx
A 0
B −1
C ∞
D 1
sinxtanx=sinxsinx/cosx=cosxtanxsinx=sinx/cosxsinx=cosx. As x→0, cos0=1cos0=1. Hence the limit is 1.
Evaluate limx→0tanx−sinxx3limx→0x3tanx−sinx
A 0
B 1/3
C 1/2
D 1
Use expansions: tanx=x+x33+…tanx=x+3×3+… and sinx=x−x36+…sinx=x−6×3+…. Difference ≈x33+x36=x32≈3×3+6×3=2×3. So limit is 1/2.
Evaluate limx→01−cosxsin2xlimx→0sin2x1−cosx
A 1/2
B 1
C 2
D 0
Use 1−cosx∼x221−cosx∼2×2 and sinx∼xsinx∼x, so sin2x∼x2sin2x∼x2. Ratio ∼x2/2×2=1/2∼x2x2/2=1/2.
If f and g are continuous at a, then f(g(x)) is continuous when
A g(a) exists
B f at g(a)
C both A and B
D none
For composite continuity at a, g must be continuous at a so g(a) is meaningful, and f must be continuous at g(a). Then f(g(x)) is continuous at a.
For continuity at x=a, which is NOT required
A f(a) exists
B f′(a) exists
C limit exists
D limit equals f(a)
Continuity does not require differentiability. Only existence of f(a), existence of limx→af(x)limx→af(x), and equality of the two are needed for continuity at a.
A function with jump at x=a is always
A discontinuous
B continuous
C differentiable
D constant
Jump means left and right limits differ, so the two-sided limit does not exist. Without a single limit value, continuity at that point is impossible.
If limx→af(x)=0limx→af(x)=0 and f(a)=0, then at a
A discontinuous
B infinite break
C continuous
D oscillatory
Continuity requires that the limit exists and equals the function value. Since both are 0 here, and f(a) exists, the function is continuous at x=a.
A function can be discontinuous but have derivative at a
A Yes always
B Only if jump
C Only if hole
D Never possible
Differentiability implies continuity. If a function is discontinuous at a point, the derivative cannot exist there. So a discontinuous function cannot be differentiable at that point.
Check differentiability of ∣x∣∣x∣ at x=0 using slopes
A slopes −1 and 1
B both slopes 1
C both slopes 0
D slopes infinite
For x<0, |x|=−x so slope is −1; for x>0, |x|=x so slope is 1. Unequal one-sided slopes mean derivative at 0 does not exist.
The derivative at x=a equals
A secant slope
B average value
C tangent slope
D area value
The derivative at a point is the limit of slopes of secant lines as the second point approaches the first. This limit gives the slope of the tangent line.
For f(x)=x2f(x)=x2, derivative at x=3 is
A 3
B 6
C 9
D 12
f′(x)=2xf′(x)=2x for x2x2. Substituting x=3 gives f′(3)=6f′(3)=6. This is the instantaneous rate of change of x² at x=3.
For f(x)=sinxf(x)=sinx, f′(0) equals
A 0
B −1
C 1
D ∞
The derivative of sin x is cos x. Evaluating at 0 gives cos 0 = 1. So the slope of y=sin x at x=0 is 1.
For f(x)=cosxf(x)=cosx, f′(0) equals
A 1
B −1
C 2
D 0
The derivative of cos x is −sin x. At x=0, sin 0 = 0, so f′(0)=−0=0. The tangent is horizontal there.
If y = e^x, then y″ equals
A e^x
B 0
C −e^x
D x e^x
Derivative of e^x is e^x. Differentiating again gives e^x again. So all higher derivatives of e^x are e^x.
If y = ln x, then y′ equals
A x
B ln x
C 1/x
D 0
The derivative of ln x is 1/x for x>0. It follows from inverse relationship with e^x and can also be derived using limits.
If y = x3x3, then y″ at x=1 equals
A 3
B 6
C 0
D 12
y′=3x² and y″=6x. At x=1, y″=6. The second derivative indicates concavity and how fast slope changes.
Using chain rule, derivative of sin(2x)sin(2x) is
A cosxcosx
B sin2xsin2x
C 2cos2x2cos2x
D 2sin2x2sin2x
For sin(u)sin(u), derivative is cos(u)⋅u′cos(u)⋅u′. Here u=2x so u′=2. Thus derivative is 2cos(2x)2cos(2x).
Derivative of (x2+1)5(x2+1)5 is
A 10x(x2+1)410x(x2+1)4
B 5(x2+1)45(x2+1)4
C 2x(x2+1)52x(x2+1)5
D 10(x2+1)410(x2+1)4
Use chain rule: derivative of (u^5) is 5u^4·u′. Here u=x²+1, u′=2x. So result is 5(x2+1)4⋅2×5(x2+1)4⋅2x.
Derivative of 1xx1 is
A 1/x
B 1/x²
C −x²
D −1/x²
Write 1/x as x^{-1}. Differentiate: ddxx−1=−1⋅x−2=−1/x2dxdx−1=−1⋅x−2=−1/x2. Valid for x≠0.
Evaluate limx→1lnxx−1limx→1x−1lnx
A 0
B e
C 1
D ∞
This is a standard limit: limx→1lnxx−1=1limx→1x−1lnx=1. It can be shown by substitution x=1+h and using ln(1+h)∼hln(1+h)∼h.
Evaluate limx→0ax−1xlimx→0xax−1 (a>0)
A a
B ln a
C 1/a
D 0
A standard result is limx→0ax−1x=lnalimx→0xax−1=lna. It comes from writing ax=exlnaax=exlna and using the e^u limit.
For indeterminate form 0^0, a common method is
A direct substitute
B always 0
C take logarithm
D always 1
For forms like f(x)g(x)f(x)g(x) giving 0^0 or 1^∞, take logs: ln y = g(x) ln f(x). Convert to product/quotient and evaluate limit, then exponentiate back.
If limx→af(x)=Llimx→af(x)=L and f is continuous at a, then
A f(a)=L
B f(a)≠L
C LHL≠RHL
D f(a) undefined
Continuity at a directly means the function value equals the limit. So if limit is L and the function is continuous, f(a) must be L.
If f(x) has vertical tangent at x=a, then
A derivative zero
B function discontinuous
C derivative infinite
D limit absent
A vertical tangent means the slope becomes unbounded as you approach the point. So the derivative tends to ±∞, and the usual finite derivative does not exist.
For y=x2exy=x2ex, which rule is needed first
A quotient rule
B inverse rule
C constant rule
D product rule
The function is a product of x² and e^x. Apply product rule: (x²)’e^x + x²(e^x)’. Then simplify, and you may also factor common terms.
Leibnitz theorem gives nth derivative of
A sum of two
B product of two
C difference of two
D inverse function
Leibnitz’s theorem generalizes product rule for higher derivatives: (fg)(n)=∑k=0n(nk)f(k)g(n−k)(fg)(n)=∑k=0n(kn)f(k)g(n−k). It’s widely used in nth derivative problems.
In Leibnitz formula, total terms are
A n
B 2n
C n+1
D 2^n
The summation runs from k=0 to k=n, which includes n+1 values. Hence there are n+1 terms in the Leibnitz expansion for the nth derivative of a product.
For f(x)=xnf(x)=xn and g(x)=eaxg(x)=eax, nth derivative of gg is
A aneaxaneax
B aeaxaeax
C naeaxnaeax
D exex
Each differentiation of eaxeax brings a factor a. So the nth derivative is aneaxaneax. This is useful inside Leibnitz theorem for products.
Evaluate limx→0sinx−xx3limx→0x3sinx−x
A 0
B 1/6
C −1/6
D −1/3
Use expansion sinx=x−x36+…sinx=x−6×3+…. Then sinx−x≈−x36sinx−x≈−6×3. Dividing by x3x3 gives −1/6.
Evaluate limx→01−cosxx2limx→0x21−cosx
A 1
B 2
C 0
D 1/2
Near 0, cosx=1−x22+…cosx=1−2×2+…. So 1−cosx≈x221−cosx≈2×2. Dividing by x2x2 yields 1/2.
Evaluate limx→0sin2xx2limx→0x2sin2x
A 1
B 0
C 2
D 1/2
sin2xx2=(sinxx)2x2sin2x=(xsinx)2. Since sinxx→1xsinx→1 as x→0, the square also tends to 1.
If f′(a)=0f′(a)=0, then tangent at x=a is
A vertical
B horizontal
C undefined
D curved only
The derivative gives the slope of the tangent. If f′(a)=0f′(a)=0, slope is zero, meaning the tangent line is horizontal at that point.
If f is continuous on [a,b], then it satisfies
A L’Hospital
B Leibnitz only
C IVT always
D Discontinuous theorem
The Intermediate Value Theorem requires continuity on [a,b]. It guarantees f takes every value between f(a) and f(b), which is a key property of continuous graphs.
A discontinuity at endpoint can be checked using
A one-sided limit
B two-sided limit
C derivative only
D second derivative
At endpoints, only one side exists within the interval. So continuity is checked using right-hand limit at left endpoint and left-hand limit at right endpoint.
If f(x)=∣x∣xf(x)=x∣x∣ for x≠0, then near 0 it is
A continuous
B removable
C jump discontinuous
D oscillatory
For x>0, |x|/x=1; for x<0, |x|/x=−1. Left and right limits at 0 are −1 and 1, not equal, so jump discontinuity.
If limx→af(x)=∞limx→af(x)=∞, then at x=a there is
A removable break
B no break
C corner
D infinite break
When function values blow up without bound near a point, the graph has a vertical asymptote. This is called infinite discontinuity.
If two functions squeeze f(x) and both limits are 0, then
A f limit 1
B f limit 0
C f diverges
D f oscillates
By squeeze theorem, if g(x)≤f(x)≤h(x)g(x)≤f(x)≤h(x) near a and limg=limh=0limg=limh=0, then f must also approach 0.
Evaluate limx→0x−sinxx3limx→0x3x−sinx
A −1/6
B 0
C 1/6
D 1/3
Using sinx=x−x36+…sinx=x−6×3+…, we get x−sinx≈x36x−sinx≈6×3. Dividing by x3x3 gives 1/6.
For y=ln(sinx)y=ln(sinx), derivative requires
A chain rule
B quotient rule
C Leibnitz rule
D no rule
Differentiate ln(u) as u′/u. Here u=sin x, u′=cos x. So derivative is cosxsinx=cotxsinxcosx=cotx. It uses chain rule clearly.
If f(x)f(x) is continuous at a and g(x) → a, then limf(g(x))limf(g(x)) equals
A g(a)
B 0
C f(a)
D ∞
This is a key continuity property: if g(x) approaches a and f is continuous at a, then f(g(x)) approaches f(a). It links limits with continuity.
For y=x2sinxy=x2sinx, second derivative needs
A only power rule
B only quotient rule
C only limits
D product rule
First derivative already needs product rule for x² and sin x. Second derivative differentiates that result again, using product rule again and basic trig derivatives.