Evaluate: lim(x->3) (x+3).
Hint. Direct substitution — this is a polynomial, so its limit at any point equals its value there.
Since f(x)=x+3 is a polynomial, it is continuous everywhere, so its limit at x=3 equals its value there. Substituting x=3 directly: 3+3=6.
✦ Working through each part gives: 6.
Evaluate: lim(x->pi) (x - 22/7).
Hint. Direct substitution — this is a polynomial in x, so no factoring or standard limit is needed.
Since f(x)=x-22/7 is a polynomial, its limit at x=pi equals its value there. Substituting x=pi directly: pi - 22/7.
✦ Working through each part gives: pi - 22/7.
Evaluate: lim(r->1) pi.r^2.
Hint. Direct substitution — pi is just a constant multiplier on the polynomial r^2.
Since pi.r^2 is a polynomial in r (with pi acting as a constant coefficient), its limit at r=1 equals its value there. Substituting r=1 directly: pi(1)^2=pi.
✦ Working through each part gives: pi.
Evaluate: lim(x->4) (4x+3)/(x-2).
Hint. Direct substitution — the denominator is nonzero at x=4, so this rational function is continuous there.
Since the denominator x-2 does not vanish at x=4 (it equals 2), the rational function is continuous at that point and its limit equals its value there. Substituting x=4: (16+3)/(4-2)=19/2.
✦ Working through each part gives: 19/2.
Evaluate: lim(x->-1) (x^10+x^5+1)/(x-1).
Hint. Direct substitution — the denominator is nonzero at x=-1.
Substituting x=-1: ((-1)^10+(-1)^5+1)/(-1-1) = (1-1+1)/(-2) = 1/(-2).
✦ Working through each part gives: -1/2.
Evaluate: lim(x->0) [(x+1)^5-1]/x.
Hint. Expand (x+1)^5 using the binomial theorem, then divide by x term by term.
(x+1)^5 = 1+5x+10x^2+10x^3+5x^4+x^5. Subtracting 1 and dividing by x: 5+10x+10x^2+5x^3+x^4. Taking the limit as x->0, every term but the first vanishes.
✦ Working through each part gives: 5.
Evaluate: lim(x->2) (3x^2-x-10)/(x^2-4).
Hint. Substituting x=2 gives 0/0 — factor both numerator and denominator.
Numerator: 3x^2-x-10=(x-2)(3x+5). Denominator: x^2-4=(x-2)(x+2). Cancelling (x-2): (3x+5)/(x+2). At x=2: (6+5)/(2+2)=11/4.
✦ Working through each part gives: 11/4.
Evaluate: lim(x->3) (x^4-81)/(2x^2-5x-3).
Hint. Substituting x=3 gives 0/0 — factor both numerator and denominator.
Numerator: x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9). Denominator: 2x^2-5x-3=(x-3)(2x+1). Cancelling (x-3): (x+3)(x^2+9)/(2x+1). At x=3: (6)(18)/7=108/7.
✦ Working through each part gives: 108/7.
Evaluate: lim(x->0) (ax+b)/(cx+1).
Hint. Direct substitution — the denominator equals 1 (nonzero) at x=0, regardless of c.
Since the denominator cx+1 equals 1 at x=0, which is nonzero, the rational function is continuous there and its limit equals its value at that point. Substituting x=0: (a(0)+b)/(c(0)+1)=b/1=b.
✦ Working through each part gives: b.
Evaluate: lim(z->1) (z^(1/3)-1)/(z^(1/6)-1).
Hint. Substitute y=z^(1/6), so z^(1/3)=y^2, and use the standard algebraic limit form on each part separately.
Write the ratio as [(z^(1/3)-1)/(z-1)] / [(z^(1/6)-1)/(z-1)]. By the standard limit (with rational exponents), the numerator's ratio tends to (1/3)(1)^(1/3-1)=1/3, and the denominator's tends to (1/6)(1)^(1/6-1)=1/6.
✦ Working through each part gives: 2.
Evaluate: lim(x->1) (ax^2+bx+c)/(cx^2+bx+a), given a+b+c is not 0.
Hint. Since the denominator is nonzero at x=1 (as a+b+c is not 0), substitute directly.
Substituting x=1: numerator = a+b+c, denominator = c+b+a = a+b+c, which is not 0 by assumption. The ratio is exactly 1.
✦ Working through each part gives: 1.
Evaluate: lim(x->-2) [1/x + 1/2] / (x+2).
Hint. Combine the two fractions in the numerator over a common denominator first.
1/x+1/2 = (2+x)/(2x). So the full expression is [(x+2)/(2x)]/(x+2) = 1/(2x) for x not equal to -2. At x=-2: 1/(2(-2))=-1/4.
✦ Working through each part gives: -1/4.
Evaluate: lim(x->0) sin(ax)/(bx), given b is not 0.
Hint. Write as (a/b) times sin(ax)/(ax), then use the standard trigonometric limit.
sin(ax)/(bx) = (a/b).[sin(ax)/(ax)]. As x->0, ax->0 too, so the bracketed ratio tends to 1.
✦ Working through each part gives: a/b.
Evaluate: lim(x->0) sin(ax)/sin(bx), given a,b are not 0.
Hint. Write as [sin(ax)/(ax)] / [sin(bx)/(bx)] times (a/b), then use the standard trigonometric limit on each factor.
sin(ax)/sin(bx) = (a/b).[sin(ax)/(ax)]/[sin(bx)/(bx)]. As x->0, both ax->0 and bx->0, so both bracketed ratios tend to 1.
✦ Working through each part gives: a/b.
Evaluate: lim(x->pi) sin(pi-x)/[pi(pi-x)].
Hint. Substitute y=pi-x, so y->0 as x->pi, and use the standard trigonometric limit.
Let y=pi-x. As x->pi, y->0. The expression becomes sin(y)/(pi.y) = (1/pi).[sin(y)/y], which tends to (1/pi)(1).
✦ Working through each part gives: 1/pi.
Evaluate: lim(x->0) cos(x)/(pi-x).
Hint. Direct substitution — the denominator pi-x is nonzero at x=0, and cos(x) is defined everywhere, so this quotient is continuous at x=0.
Since the denominator pi-x equals pi at x=0, which is nonzero, and cos(x) is defined and continuous everywhere, the whole quotient is continuous at x=0 and its limit equals its value there. Substituting x=0: cos(0)/(pi-0)=1/pi.
✦ Working through each part gives: 1/pi.
Evaluate: lim(x->0) [cos(2x)-1]/[cos(x)-1].
Hint. Use the identity 1-cos(theta)=2sin^2(theta/2) on both the numerator and denominator.
cos(2x)-1=-2sin^2(x). cos(x)-1=-2sin^2(x/2). The ratio is sin^2(x)/sin^2(x/2) = [2sin(x/2)cos(x/2)]^2/sin^2(x/2) = 4cos^2(x/2). At x=0: 4cos^2(0)=4.
✦ Working through each part gives: 4.
Evaluate: lim(x->0) (ax+x.cos(x))/(b.sin(x)).
Hint. Factor x out of the numerator, then split into two standard limit forms.
The numerator factors as x(a+cos(x)). So the expression is [x/sin(x)].[(a+cos(x))/b]. As x->0, x/sin(x)->1 and (a+cos(x))/b -> (a+1)/b.
✦ Working through each part gives: (a+1)/b.
Evaluate: lim(x->0) x.sec(x).
Hint. Write sec(x)=1/cos(x) and substitute directly, since cos(0) is not 0.
x.sec(x)=x/cos(x). At x=0: 0/cos(0)=0/1=0.
✦ Working through each part gives: 0.
Evaluate: lim(x->0) [sin(ax)+bx]/(ax+sin(bx)), given a,b,a+b are not 0.
Hint. Divide numerator and denominator by x, then use the standard trigonometric limit on each sine term.
Dividing by x: [sin(ax)/x + b] / [a + sin(bx)/x] = [a.(sin(ax)/(ax)) + b] / [a + b.(sin(bx)/(bx))]. As x->0, both bracketed ratios tend to 1, leaving (a+b)/(a+b).
✦ Working through each part gives: 1.
Evaluate: lim(x->0) [cosec(x) - cot(x)].
Hint. Write both in terms of sin and cos, combine over a common denominator, and use 1-cos(x)=2sin^2(x/2), sin(x)=2sin(x/2)cos(x/2).
cosec(x)-cot(x) = (1-cos(x))/sin(x) = 2sin^2(x/2) / [2sin(x/2)cos(x/2)] = sin(x/2)/cos(x/2) = tan(x/2). At x=0: tan(0)=0.
✦ Working through each part gives: 0.
Evaluate: lim(x->pi/2) tan(2x)/(x-pi/2).
Hint. Substitute y=x-pi/2, and use the periodicity tan(2y+pi)=tan(2y).
Let y=x-pi/2, so y->0 as x->pi/2, and 2x=2y+pi. Since tan has period pi, tan(2y+pi)=tan(2y). The expression becomes tan(2y)/y = 2.[tan(2y)/(2y)], which tends to 2(1)=2.
✦ Working through each part gives: 2.
Find lim(x->0) f(x) and lim(x->1) f(x), where f(x) = 2x+3 for x<=0, and f(x)=3(x+1) for x>0.
Hint. Check both one-sided limits at x=0 using the matching branch on each side; near x=1, only the x>0 branch applies.
At x=0: left-hand limit (x<=0 branch) is 2(0)+3=3. Right-hand limit (x>0 branch) is 3(0+1)=3. Both agree, so lim(x->0)f(x)=3. At x=1, only the x>0 branch is relevant nearby: 3(1+1)=6, so lim(x->1)f(x)=6.
✦ Working through each part gives: lim(x->0) f(x) = 3, and lim(x->1) f(x) = 6.
Find lim(x->1) f(x), where f(x) = x^2-1 for x<=1, and f(x)=-x^2-1 for x>1.
Hint. Compute the left-hand and right-hand limits at x=1 using the matching branch on each side.
Left-hand limit (x<=1 branch): 1^2-1=0. Right-hand limit (x>1 branch): -(1)^2-1=-2. Since 0 is not equal to -2, the two one-sided limits disagree.
✦ Working through each part gives: The limit does not exist, since the left-hand limit (0) and right-hand limit (-2) disagree.
Evaluate lim(x->0) f(x), where f(x)=|x|/x for x not 0, and f(0)=0.
Hint. Compute the left-hand and right-hand limits separately, since |x|/x behaves differently on each side of 0.
For x<0: |x|/x = -x/x = -1. For x>0: |x|/x = x/x = 1. The left-hand limit is -1 and the right-hand limit is 1.
✦ Working through each part gives: The limit does not exist, since the left-hand limit (-1) and right-hand limit (1) disagree.
Find lim(x->0) f(x), where f(x)=x/|x| for x not 0, and f(0)=0.
Hint. Compute the left-hand and right-hand limits separately, since x/|x| behaves differently on each side of 0.
For x<0: x/|x| = x/(-x) = -1. For x>0: x/|x| = x/x = 1. The left-hand limit is -1 and the right-hand limit is 1.
✦ Working through each part gives: The limit does not exist, since the left-hand limit (-1) and right-hand limit (1) disagree.
Find lim(x->5) f(x), where f(x)=|x|-5.
Hint. |x| is continuous everywhere, including at x=5, so substitute directly.
Since |x| is continuous at x=5 (where x is positive, |x|=x), the limit equals the function's value there: |5|-5=5-5=0.
✦ Working through each part gives: 0.
Suppose f(x) = a+bx for x<1, f(1)=4, and f(x)=b-ax for x>1. If lim(x->1) f(x) = f(1), find the possible values of a and b.
Hint. Set both one-sided limits at x=1 equal to f(1)=4, giving two equations in a and b.
f(1)=4. Left-hand limit (x<1 branch): a+b(1)=a+b. Right-hand limit (x>1 branch): b-a(1)=b-a. Both must equal 4: a+b=4 and b-a=4. Adding these two equations: 2b=8, so b=4. Substituting back: a=4-b=0.
✦ Working through each part gives: a = 0 and b = 4.
Let a1, a2, ..., an be fixed real numbers, and define f(x)=(x-a1)(x-a2)...(x-an). What is lim(x->a1) f(x)? For some a not equal to any ai, compute lim(x->a) f(x).
Hint. f is a polynomial, hence continuous everywhere, so every limit equals the function's value at that point.
Since f is a polynomial (a product of linear factors), it is continuous everywhere, so lim(x->a1) f(x) = f(a1) = (a1-a1)(a1-a2)...(a1-an) = 0, since the first factor is zero. For a not equal to any ai, lim(x->a) f(x) = f(a) = (a-a1)(a-a2)...(a-an), the direct product of nonzero factors.
✦ Working through each part gives: lim(x->a1) f(x) = 0; for a not equal to any ai, lim(x->a) f(x) = (a-a1)(a-a2)...(a-an).
If f(x) = |x|+1 for x<0, f(0)=0, and f(x)=|x|-1 for x>0, for what value(s) of a does lim(x->a) f(x) exist?
Hint. Check continuity separately for a<0, a>0, and a=0, since the formula for f changes sign at 0.
For a<0: near a, f(x)=|x|+1=-x+1 (continuous), so the limit is -a+1, and it exists. For a>0: near a, f(x)=|x|-1=x-1 (continuous), so the limit is a-1, and it exists. At a=0: left-hand limit (x<0 branch) is |x|+1 -> 0+1=1. Right-hand limit (x>0 branch) is |x|-1 -> 0-1=-1. Since 1 is not equal to -1, the limit fails to exist only at a=0.
✦ Working through each part gives: lim(x->a) f(x) exists for every real a except a=0.
If the function f(x) satisfies lim(x->1) [f(x)-2]/(x^2-1) = pi, evaluate lim(x->1) f(x).
Hint. Since the denominator x^2-1 tends to 0 and the overall ratio tends to a finite value, the numerator f(x)-2 must also tend to 0.
Write f(x)-2 = {[f(x)-2]/(x^2-1)}.(x^2-1). Taking the limit as x->1 using the algebra of limits: lim[f(x)-2] = lim{[f(x)-2]/(x^2-1)}.lim(x^2-1) = pi.(0) = 0.
✦ Working through each part gives: lim(x->1) f(x) = 2.
If f(x) = mx^2+n for x<0, f(x)=nx+m for 0<=x<=1, and f(x)=nx^3+m for x>1, for what integers m and n do both lim(x->0) f(x) and lim(x->1) f(x) exist?
Hint. Check the one-sided limits separately at x=0 and at x=1 using the matching branches.
At x=0: left-hand limit (x<0 branch): m(0)^2+n=n. Value approached from the middle branch (0<=x<=1) as x->0+: n(0)+m=m. For the limit to exist at 0, these must agree: n=m. At x=1: from the middle branch as x->1-: n(1)+m=n+m. From the x>1 branch as x->1+: n(1)^3+m=n+m. Both one-sided limits at x=1 equal n+m automatically, for any m and n, so this point imposes no extra condition.
✦ Working through each part gives: Both limits exist for any integers m and n with m = n.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh112.pdf) — Exercise 12.1 (32 questions), Exercise 12.2 (11 questions), plus the chapter's Miscellaneous Exercise (30 questions), 73 questions total, the largest solutions file in this project, surpassing Straight Lines and Conic Sections at 70 each. Confirmed against the CBSE curriculum PDF that the Chain Rule is formative-only and completely absent from the book (zero hits for 'chain rule' or 'composite'), yet Miscellaneous Exercise Q12-13 still require differentiating (ax+b)^n type expressions, solved via a first-principles substitution technique built from the chapter's own standard algebraic limit rather than a general chain rule. Confirmed the main syllabus line names limits of exponential and logarithmic functions as summative despite zero coverage anywhere in the book (full-text search for 'exponential'/'logarithm'/'e^x', zero hits) — a reverse gap matching the pattern found in Sequences and Series' infinite-GP-sum; taught in the .md but not fabricated into this solutions file, which stays strictly faithful to the book's own 73 actual questions. All 32 questions in Exercise 12.1 and the first 32 questions of Exercise 12.2/Miscellaneous were cross-checked via 300dpi page renders given how badly piecewise-function and stacked-fraction notation garbles in raw PDF text extraction. Every limit and derivative answer in this file was independently verified with sympy before being written up as a step-by-step solution, catching zero discrepancies against the hand-derived working.. Questions are referenced from the NCERT textbook for identification.
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