Exercise 12.1 answers: Limits and Derivatives

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1. x + 3 has no fraction and no break in it — it is a straight, unbroken line.
2. So put x = 3 straight in: 3 + 3.

Answer6

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1. x − 22/7 is a polynomial in x (a straight-line rule with no x in a denominator), so its limit is found by putting the value in.
2. Put x = π: π − 22/7.
3. π is not exactly 22/7 (22/7 is only an approximation), so this is not 0; it is left as π − 22/7.

Answerπ − 22/7

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1. π is just a fixed number multiplying r², and r² has no break anywhere.
2. So put r = 1 straight in: π × 1² = π.

Answerπ

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1. Check the bottom at x = 4 first: 4 − 2 = 2, which is not zero.
2. Since the bottom survives, put x = 4 into the top too: 4×4 + 3 = 19.
3. The limit is 19/2.

Answer19/2

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1. Check the bottom at x = −1 first: −1 − 1 = −2, not zero.
2. Put x = −1 into the top: (−1)¹⁰ + (−1)⁵ + 1 = 1 − 1 + 1 = 1.
3. The limit is 1 ÷ (−2) = −1/2.

Answer−1/2

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1. Both the top and the bottom become 0 at x = 0, so we cannot substitute yet.
2. Expand (x + 1)⁵ using the binomial pattern: 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵.
3. Subtract the 1: top becomes 5x + 10x² + 10x³ + 5x⁴ + x⁵.
4. Divide every term by x: 5 + 10x + 10x² + 5x³ + x⁴, and now x = 0 is safe.
5. Put x = 0: only the 5 is left.

Answer5

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1. Both top and bottom become 0 at x = 2, so first pull out the common factor.
2. Top factors as (3x + 5)(x − 2); bottom factors as (x − 2)(x + 2).
3. Cancel the shared (x − 2): what is left is (3x + 5)/(x + 2).
4. Put x = 2: 11/4.

Answer11/4

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1. Both top and bottom become 0 at x = 3, so pull out the common factor.
2. Top factors as (x − 3)(x + 3)(x² + 9); bottom factors as (x − 3)(2x + 1).
3. Cancel the shared (x − 3): what is left is (x + 3)(x² + 9)/(2x + 1).
4. Put x = 3: (6 × 18)/7 = 108/7.

Answer108/7

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1. Evaluate: lim (ax + b)/(cx + 1) (x → 0).
2. At x = 0 the bottom is c×0 + 1 = 1, which is never zero, whatever c is.
3. So substitute x = 0 into the top too: a×0 + b = b.
4. The limit is b ÷ 1 = b.

Answerb

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1. Both top and bottom become 0 at z = 1, and neither is a plain polynomial.
2. Let y = z^(1/6), so as z → 1, y → 1 too, and z^(1/3) = y², z^(1/6) = y.
3. The rule becomes (y² − 1)/(y − 1) = (y − 1)(y + 1)/(y − 1) = y + 1, after cancelling.
4. Put y = 1: 1 + 1 = 2.

Answer2

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1. Evaluate: lim (ax² + bx + c)/(cx² + bx + a) (x → 1), a + b + c ≠ 0.
2. Put x = 1 into the top: a + b + c.
3. Put x = 1 into the bottom: c + b + a, the very same sum.
4. Since we are told a + b + c ≠ 0, the bottom is safe, so the limit is (a+b+c)/(a+b+c) = 1.

Answer1

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1. First combine the top into one fraction: 1/x + 1/2 = (2 + x)/(2x).
2. So the whole rule is (2 + x)/(2x) ÷ (x + 2) = (2 + x)/(2x(x + 2)).
3. Cancel the shared (x + 2): what is left is 1/(2x).
4. Put x = −2: 1/(2 × −2) = −1/4.

Answer−1/4

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1. Evaluate: lim sin ax/bx (x → 0).
2. Write sin(ax)/(bx) as (a/b) × sin(ax)/(ax).
3. As x → 0, ax → 0 too, and the standard result sin(ax)/(ax) → 1 applies.
4. So the limit is (a/b) × 1 = a/b.

Answera/b

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1. Evaluate: lim sin ax/sin bx (x → 0), a, b ≠ 0.
2. Split the rule into three pieces: sin(ax)/(ax), then (ax)/(bx) = a/b, then bx/sin(bx).
3. As x → 0, both sin(ax)/(ax) and bx/sin(bx) go to 1.
4. So the limit is 1 × (a/b) × 1 = a/b.

Answera/b

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1. Evaluate: lim sin(π − x)/[π(π − x)] (x → π).
2. Let t = π − x, so as x → π, t → 0.
3. The rule becomes sin(t)/(π t) = (1/π) × sin(t)/t.
4. As t → 0, sin(t)/t → 1, so the limit is 1/π.

Answer1/π

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1. Check the bottom at x = 0: π − 0 = π, which is not zero.
2. Put x = 0 into the top: cos 0 = 1.
3. The limit is 1/π.

Answer1/π

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1. Evaluate: lim (cos 2x − 1)/(cos x − 1) (x → 0).
2. Both top and bottom go to 0 at x = 0, so use the identity 1 − cos θ = 2 sin²(θ/2).
3. cos 2x − 1 = −2 sin²x, and cos x − 1 = −2 sin²(x/2).
4. The rule becomes sin²x / sin²(x/2); write sin x = 2 sin(x/2)cos(x/2) to get 4 cos²(x/2).
5. Put x = 0: 4 × 1² = 4.

Answer4

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1. Both top and bottom go to 0 at x = 0, so factor x out of the top first.
2. Top becomes x(a + cos x), so the rule is [(a + cos x)/b] × [x/sin x].
3. As x → 0: a + cos x → a + 1, and x/sin x → 1.
4. So the limit is (a + 1)/b.

Answer(a + 1)/b

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1. Evaluate: lim x sec x (x → 0).
2. Write sec x as 1/cos x, so x sec x = x/cos x.
3. At x = 0 the bottom is cos 0 = 1, which is not zero, so we may put x = 0 straight in.
4. The top is 0 and the bottom is 1, so the limit is 0/1 = 0.

Answer0

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1. Evaluate: lim (x→0) (sin ax+bx)/(ax+sin bx) a,b, a+b≠0.
2. As x → 0, both sin ax and sin bx behave like ax and bx (since sin θ/θ → 1 when θ → 0).
3. So the top, sin ax + bx, behaves like ax + bx.
4. And the bottom, ax + sin bx, also behaves like ax + bx.
5. Divide the whole fraction by x: top becomes (sin ax)/x + b = a·(sin ax)/(ax) + b → a·1 + b = a + b.
6. Bottom becomes a + (sin bx)/x = a + b·(sin bx)/(bx) → a + b·1 = a + b.
7. So the limit is (a + b)/(a + b), and since a + b ≠ 0 this is 1.

Answer1

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1. Evaluate: lim (x→0) (cosec x − cot x).
2. Write cosec x as 1/sin x and cot x as cos x/sin x.
3. So cosec x − cot x = (1 − cos x)/sin x.
4. As x → 0, the top 1 − cos x → 0 and the bottom sin x → 0 too, so simplify further.
5. Use 1 − cos x = 2 sin²(x/2) and sin x = 2 sin(x/2) cos(x/2).
6. The fraction becomes sin(x/2)/cos(x/2) = tan(x/2), which → tan(0) = 0.

Answer0

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1. Put h = x − π/2, so x = π/2 + h, and h → 0 as x → π/2.
2. tan 2x = tan(π + 2h) = tan 2h, because tan repeats every π.
3. So the limit becomes lim (h→0) (tan 2h)/h.
4. Multiply top and bottom by 2: 2·(tan 2h)/(2h), and (tan θ)/θ → 1 as θ → 0.
5. So the limit is 2 × 1 = 2.

Answer2

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1. Given: f(x) = 2x + 3 for x ≤ 0 and 3(x + 1) for x > 0. Find the limits of f(x) as x → 0 and as x → 1.
2. Near x = 0 from the left, f(x) = 2x + 3, which → 2(0) + 3 = 3.
3. Near x = 0 from the right, f(x) = 3(x + 1), which → 3(0 + 1) = 3.
4. Both sides give 3, so lim (x→0) f(x) = 3.
5. x = 1 is greater than 0, so near x = 1 both sides use f(x) = 3(x + 1).
6. 3(x + 1) → 3(1 + 1) = 6, so lim (x→1) f(x) = 6.

Answerlim (x→0) f(x) = 3; lim (x→1) f(x) = 6

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1. Given: f(x) = x² − 1 for x ≤ 1 and −x² − 1 for x > 1. Find the limit of f(x) as x → 1.
2. Near x = 1 from the left, f(x) = x² − 1, which → 1² − 1 = 0.
3. Near x = 1 from the right, f(x) = −x² − 1, which → −(1)² − 1 = −2.
4. The left value (0) and the right value (−2) are different.
5. So the limit does not exist at x = 1.

Answerdoes not exist

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1. Given: f(x) = |x|/x for x ≠ 0 and 0 for x = 0. Evaluate the limit of f(x) as x → 0.
2. For x < 0, |x| = −x, so |x|/x = −x/x = −1. So the left-hand limit is −1.
3. For x > 0, |x| = x, so |x|/x = x/x = 1. So the right-hand limit is 1.
4. The left-hand limit (−1) and the right-hand limit (1) are different.
5. So lim (x→0) f(x) does not exist.

Answerdoes not exist

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1. Given: f(x) = x/|x| for x ≠ 0 and 0 for x = 0. Find the limit of f(x) as x → 0.
2. For x < 0, |x| = −x, so x/|x| = x/(−x) = −1. So the left-hand limit is −1.
3. For x > 0, |x| = x, so x/|x| = x/x = 1. So the right-hand limit is 1.
4. The left-hand limit (−1) and the right-hand limit (1) are different.
5. So lim (x→0) f(x) does not exist.

Answerdoes not exist

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1. For x near 5, x is positive, so |x| = x.
2. So near x = 5, f(x) is just x − 5, an ordinary continuous function.
3. Put x = 5 directly: f(5) = |5| − 5 = 5 − 5 = 0.
4. So lim (x→5) f(x) = 0.

Answer0

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1. Given: f(x) = a + bx for x < 1, 4 for x = 1, b − ax for x > 1, and the limit of f(x) as x → 1 equals f(1).
2. f(1) = 4, so both one-sided limits at x = 1 must also equal 4.
3. From the left, f(x) = a + bx → a + b as x → 1. So a + b = 4.
4. From the right, f(x) = b − ax → b − a as x → 1. So b − a = 4.
5. Add the two equations: (a + b) + (b − a) = 8, so 2b = 8, giving b = 4.
6. Put b = 4 into a + b = 4: a = 0.

Answera = 0, b = 4

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1. f(x) is a polynomial (a product of simple factors), so it is continuous everywhere.
2. For a continuous function, the limit at any point equals the function's value there.
3. So lim (x→a1) f(x) = f(a1) = (a1 − a1)(a1 − a2)...(a1 − an).
4. One factor, (a1 − a1), is 0, so the whole product is 0.
5. For a general a not equal to any ai, lim (x→a) f(x) = f(a) = (a − a1)(a − a2)...(a − an).

Answerlim (x→a1) f(x) = 0; lim (x→a) f(x) = (a − a1)(a − a2)...(a − an)

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1. Given: f(x) = |x| + 1 for x < 0, 0 for x = 0, |x| − 1 for x > 0. For which a does the limit of f(x) as x → a exist?
2. For a < 0, near a the function is just |x| + 1 = −x + 1, an ordinary continuous expression, so the limit exists there.
3. For a > 0, near a the function is just |x| − 1 = x − 1, also continuous, so the limit exists there.
4. At a = 0: from the left, |x| + 1 → 0 + 1 = 1. From the right, |x| − 1 → 0 − 1 = −1.
5. 1 and −1 are different, so lim (x→0) f(x) does not exist.
6. So the limit exists for every real a except a = 0.

Answerevery real number a except a = 0

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1. As x → 1, the bottom x² − 1 goes to 0.
2. If the top f(x) − 2 went to some number other than 0, the fraction would blow up and have no finite limit.
3. But we are told the fraction does have a finite limit, π.
4. So the top must also go to 0: lim (x→1) (f(x) − 2) = 0.
5. That means lim (x→1) f(x) = 2.

Answer2

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Open NCERT p. 239Checked by computerAnswers can differ: one example

1. Given: f(x) = mx² + n for x < 0, nx + m for 0 ≤ x ≤ 1, nx³ + m for x > 1. For which integers m and n do the limits as x → 0 and x → 1 both exist?
2. At x = 0: from the left, mx² + n → n. From the right (middle piece), nx + m → m. These must be equal, so m = n.
3. At x = 1: from the middle piece, nx + m → n + m. From the right, nx³ + m → n + m. These are always equal, whatever m and n are.
4. So the only condition needed is m = n, and m, n can be any pair of equal integers.
5. For example, m = 1, n = 1 works: check both limits at x = 0 and x = 1 for this pair.

AnswerBoth limits exist exactly when m = n (any integer); for example m = 1, n = 1.

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