13 सीमाएं और डेरिवेटिव

अभ्यास 1

Question:

If the functionf(x) satisfies lim(x→1) (f(x)−2)/(x^2−1)=π evaluate lim(x→1)f(x)

Answer:

Step 1: Substitute x = 1 in the equation lim(x→1) (f(x)−2)/(x^2−1)=π

lim(x→1) (f(1)−2)/(1^2−1)=π

Step 2: Solve the equation lim(x→1) (f(1)−2)/(1^2−1)=π

f(1)−2=π

f(1)=2+π

Step 3: Substitute x = 1 in the original equation lim(x→1)f(x) = lim(x→1)(2+π) = 2+π

Question:

Evaluate the given limit: lim(x→0)xsecx

Answer:

Step 1: Rewrite the expression using the definition of secant: lim(x→0)xsecx = lim(x→0)x(1/cosx)

Step 2: Rewrite the expression using the limit definition of cosine: lim(x→0)x(1/cosx) = lim(x→0)x(1/lim(x→0)cosx)

Step 3: Evaluate the limit: lim(x→0)x(1/lim(x→0)cosx) = 0/1 = 0

Question:

Evaluate the limit: lim(x→0)(sin ax+bx)/(ax+sin bx) a,b,a+b≠0

Answer:

  1. Split the fraction into two parts: lim(x→0)sin ax/ax and lim(x→0)bx/sin bx.

  2. Apply L’Hopital’s Rule to both parts: lim(x→0)a cos ax/a and lim(x→0)b/cos bx.

  3. Replace a and b with their respective values: lim(x→0)a cos ax/a = lim(x→0)cos ax and lim(x→0)b/cos bx = lim(x→0)b.

  4. The limits are equal to 1 and b, respectively.

  5. Combine the two limits: lim(x→0)(sin ax+bx)/(ax+sin bx) = 1/b.

Question:

Evaluate the Given limit: lim(r→1)πr^2

Answer:

Step 1: Substitute r=1 into the equation: lim(r→1)πr^2 = π1^2 = π

Step 2: Simplify the equation: π1^2 = π

Step 3: The answer is: π

Question:

Evaluate the given limit: lim(x→0)((x+1)^5−1)/x

Answer:

Answer: Step 1: Rewrite the limit as follows: lim(x→0)((x+1)^5−1)/x = lim(x→0)((x+1)^5)/x − lim(x→0)(1)/x

Step 2: Apply L’Hospital’s Rule to both terms: = lim(x→0)(5(x+1)^4)/1 − lim(x→0)(0)/1

Step 3: Simplify both terms: = lim(x→0)(5(x+1)^4) − lim(x→0)(0)

Step 4: Substitute 0 for x in both terms: = 5(0+1)^4 − 0

Step 5: Simplify the expression: = 5(1)^4 − 0

Step 6: Calculate the result: = 5(1) − 0

Step 7: Simplify the expression: = 5 − 0

Step 8: Calculate the result: = 5

Question:

Evaluate the Given limit: lim(x→0)sin ax/bx

Answer:

Step 1: Rewrite the limit as the quotient of two functions: lim(x→0) (sin ax)/(bx)

Step 2: Rewrite the limit using L’Hôpital’s Rule: lim(x→0) (a cos ax)/b

Step 3: Substitute 0 for x in the numerator and denominator: lim(x→0) (a cos 0)/b

Step 4: Simplify the limit: lim(x→0) a/b

Question:

Evaluate the Given limit: lim(x→π)sin(π−x)/π(π−x)

Answer:

Step 1: Rewrite the limit as lim(x→π)sin(x-π)/(x-π).

Step 2: Substitute x=π into the limit.

Step 3: lim(x→π)sin(x-π)/(x-π) = lim(x→π)sin(0)/(0) = 0/0.

Step 4: Apply L’Hospital’s Rule to the limit.

Step 5: lim(x→π)sin(x-π)/(x-π) = lim(x→π)cos(x-π)/1 = lim(x→π)cos(0) = 1.

Question:

Evaluate the Given limit: lim(x→π)(x−22/7)

Answer:

Step 1: Substitute x=π in the given expression, lim(x→π)(x−22/7) = lim(x→π)(π−22/7)

Step 2: Simplify the expression, lim(x→π)(π−22/7) = lim(x→π)(3−22/7)

Step 3: Calculate the limit, lim(x→π)(3−22/7) = 3−22/7 = -1/7

Question:

Evaluate the Given limit: lim(x→0)(ax+b)/(cx+1)

Answer:

Step 1: Substitute x=0 in the given limit.

lim(x→0)(ax+b)/(cx+1) = (a0 + b)/(c0 + 1) = b/1 = b

Step 2: The limit is equal to the value of b.

Answer: lim(x→0)(ax+b)/(cx+1) = b

Question:

Evaluate the Given limit: lim(x→0)(sin ax)/(sin bx),a,b≠0

Answer:

Step 1: Rewrite the given limit as: lim (x→0) (sin ax)/(sin bx)

Step 2: Apply L’Hôpital’s Rule: lim (x→0) (sin ax)/(sin bx) = lim (x→0) (a cos ax)/(b cos bx)

Step 3: Again, apply L’Hôpital’s Rule: lim (x→0) (a cos ax)/(b cos bx) = lim (x→0) (a^2 sin ax)/(b^2 sin bx)

Step 4: Since a and b are non-zero constants, the limit is equal to a^2/b^2.

Therefore, the answer is a^2/b^2.

Question:

Evaluate the Given limit: lim(x→−1)(x^10+x^5+1)/(x−1)

Answer:

Answer: Step 1: Rewrite the given limit in the form of fraction: lim(x→−1) (x^10+x^5+1)/(x−1) = lim(x→−1) (x^10+x^5+1)/x - lim(x→−1) (x^10+x^5+1)/1

Step 2: Apply the Limit Laws: lim(x→−1) (x^10+x^5+1)/x = -∞

Step 3: Apply the Limit Laws: lim(x→−1) (x^10+x^5+1)/1 = -1

Step 4: Subtract the two limits: lim(x→−1) (x^10+x^5+1)/(x−1) = -∞ - (-1)

Step 5: Simplify the result: lim(x→−1) (x^10+x^5+1)/(x−1) = ∞

Question:

Evaluate the given limit: lim(x→0)(cos x)/(π−x)

Answer:

  1. lim(x→0)(cos x)/(π−x)

  2. lim(x→0)cos x/(π−x)

  3. lim(x→0)cos x/π−lim(x→0)x

  4. 1/π−lim(x→0)x/π

  5. 1/π−0

  6. 1/π

Question:

Evaluate: lim(x→3)(4x^2+3)

Answer:

Step 1: Substitute x = 3 into the equation: 4(3)^2 + 3 = 39

Step 2: The limit as x approaches 3 is 39.

Question:

Evaluate the Given limit: lim(x→2) (3x^2−x−10)/(x^2−4)

Answer:

Step 1: Factor the numerator and denominator of the expression:

lim(x→2) (3x^2−x−10)/(x^2−4) = lim(x→2) (3x-5)(x+2)/(x-2)(x+2)

Step 2: Substitute x = 2 in the expression:

lim(x→2) (3x-5)(x+2)/(x-2)(x+2) = (3*2-5)(2+2)/(2-2)(2+2) = (6-5)(4)/(0)(4)

Step 3: Simplify the expression:

(6-5)(4)/(0)(4) = 4/0

Step 4: The limit does not exist as the result is undefined.

Therefore, lim(x→2) (3x^2−x−10)/(x^2−4) = undefined.

Question:

Evaluate the Given limit: lim(x→3)(x+3)

Answer:

Step 1: Substitute x=3 in the limit lim(x→3)(x+3) = 3 + 3 = 6

Step 2: The limit is equal to 6.

Question:

If f(x)=∣x∣−5 , evaluate the following limits: L(x→5)f(x)

Answer:

Answer:

Step 1: Substitute x = 5 in the given function: f(x) = |x| - 5 f(5) = |5| - 5 f(5) = 5 - 5 f(5) = 0

Step 2: Evaluate the limit: L(x→5)f(x) = 0

Question:

Evaluate: lim(x→−2)(1/x+1/2)(x+2)

Answer:

Step 1: Rewrite the expression as: lim(x→−2)(1/x)(x+2) + lim(x→−2)(1/2)(x+2)

Step 2: Substitute -2 for x in each expression: lim(x→−2)(1/x)(x+2) = 1/−2 * −2 + 2 = 0

Step 3: Evaluate the second expression: lim(x→−2)(1/2)(x+2) = 1/2 * −2 + 2 = 1

Step 4: Add the two expressions: 0 + 1 = 1

Therefore, the answer is 1.

Question:

lim(x→0)(cos 2x−1)/(cos x−1)

Answer:

Step 1: Factor out a cos x from the numerator and denominator:

lim(x→0)(cos x(cos x - 2))/(cos x - 1)

Step 2: Use the fact that cos x approaches 1 as x approaches 0:

lim(x→0)(cos x(-2))/(-1)

Step 3: Simplify:

lim(x→0)(-2cos x)/(-1)

Step 4: Factor out a -2 from the numerator and denominator:

lim(x→0)(-2(cos x))/(-2(-1))

Step 5: Simplify:

lim(x→0)(cos x)/(2)

Step 6: The answer is 1/2.

Question:

Evaluate the Given limit: lim(x→3) (x^4−81)/(2x^2−5x−3)

Answer:

Step 1: Rewrite the given expression as: lim(x→3) (x^4−81)/(2x^2−5x−3)

Step 2: Substitute x = 3 in the expression: lim(x→3) (3^4−81)/(2(3)^2−5(3)−3)

Step 3: Simplify the expression: lim(x→3) (81−81)/(18−15−3)

Step 4: Simplify the expression further: lim(x→3) 0/0

Step 5: Apply L’Hospital’s Rule: lim(x→3) (4x^3)/(4x−5)

Step 6: Substitute x = 3 in the expression: lim(x→3) (4(3)^3)/(4(3)−5)

Step 7: Simplify the expression: lim(x→3) 81/7

Step 8: The limit is 81/7.

Question:

lim(z→1) (z^(1/3)−1)/(z^(1/6)−1)

Answer:

Step 1: Simplify the expression by multiplying the numerator and denominator by (z^(1/6) + 1):

lim(z→1) (z^(1/3)−1)/(z^(1/6)−1) * (z^(1/6) + 1)

Step 2: Simplify the expression by factoring the numerator and denominator:

lim(z→1) (z^(1/2) + z^(1/3) - z^(1/6) - 1)/(z^(1/6) - 1)

Step 3: Substitute z = 1 into the expression:

lim(z→1) (1 + 1 - 1 - 1)/(1 - 1)

Step 4: Simplify the expression:

lim(z→1) 0/0

Question:

Evaluate the Given limit: lim(x→1) (ax^2+bx+c)/(cx^2+bx+a),a+b+c≠0

Answer:

Step 1: Substitute x=1 in the given expression: lim(x→1) (ax2+bx+c)/(cx2+bx+a),a+b+c≠0 = (a+b+c)/(c+b+a)

Step 2: Since a+b+c≠0, the given expression is a non-zero constant.

Step 3: Therefore, the limit of the given expression is (a+b+c)/(c+b+a).

Question:

Let a1,a2,…,an be fixed real numbers and define a function f(x)=(x−a1)(x−a2)….(x−an). What is lim(x→a1)f(x)? For some a≠a1,a2,…,an, compute lim(x→a)f(x).

Answer:

  1. lim(x→a1)f(x)=0

  2. lim(x→a)f(x)=∏(a−a1)(a−a2)….(a−an)

Question:

Evaluate the given limit: lim(x→0)(ax+x cos x)/(b sinx)

Answer:

Step 1: Rewrite the expression as

lim(x→0)(ax + xcosx)/(bsinx)

Step 2: Simplify the expression by using the properties of limits

lim(x→0)(ax + xcosx)/(bsinx) = lim(x→0)(ax + x(1 - (sinx)^2))/(bsinx)

Step 3: Apply L’Hospital’s Rule

lim(x→0)(ax + x(1 - (sinx)^2))/(bsinx) = lim(x→0)(a + x(-2sinxcosx))/(bcosx)

Step 4: Simplify the expression

lim(x→0)(a + x(-2sinxcosx))/(bcosx) = lim(x→0)(a - 2xsin^2x)/(bcosx)

Step 5: Apply L’Hospital’s Rule

lim(x→0)(a - 2xsin^2x)/(bcosx) = lim(x→0)(-2xcos^2x)/(-bsinx)

Step 6: Simplify the expression

lim(x→0)(-2xcos^2x)/(-bsinx) = lim(x→0)(2xcos^2x)/(bsinx)

Step 7: Apply L’Hospital’s Rule

lim(x→0)(2xcos^2x)/(bsinx) = lim(x→0)(2cos^2x + 2x(-2sinxcosx))/(bcosx)

Step 8: Simplify the expression

lim(x→0)(2cos^2x + 2x(-2sinxcosx))/(bcosx) = lim(x→0)(2cos^2x - 4xsin^2x)/(bcosx)

Step 9: Apply L’Hospital’s Rule

lim(x→0)(2cos^2x - 4xsin^2x)/(bcosx) = lim(x→0)(-4cos^2x - 8xsin^2xcosx)/(-bsinx)

Step 10: Simplify the expression

lim(x→0)(-4cos^2x - 8xsin^2xcosx)/(-bsinx) = lim(x→0)(4cos^2x + 8xsin^2xcosx)/(bsinx)

Step 11: Substitute x = 0

lim(x→0)(4cos^2x + 8xsin^2xcosx)/(bsinx) = 4/b

Question:

Solve: lim(x→0)(cosecx−cotx)

Answer:

Step 1: Rewrite cosecx and cotx using their definitions:

lim(x→0)(1/sinx−1/tanx)

Step 2: Use L’Hôpital’s rule:

lim(x→0)(-cosx/sinx^2−sec^2x/cosx)

Step 3: Simplify:

lim(x→0)(-1/sinx−sec^2x)

Step 4: Evaluate the limit:

lim(x→0)(-1/0−1/1)

Step 5: Answer: -1

जेईई अध्ययन सामग्री (गणित)

01 सेट

02 संबंध एवं फलन

03 त्रिकोणमितीय फलन

04 गणितीय आगमन का सिद्धांत

05 सम्मिश्र संख्याएँ और द्विघात समीकरण

06 रैखिक असमानताएँ

07 क्रमचय और संचय

08 द्विपद प्रमेय

09 अनुक्रम और श्रृंखला

10 सीधी रेखाओं का अभ्यास

10 सीधी रेखाएँ विविध

11 शांकव खंड

12 त्रिविमीय ज्यामिति का परिचय

13 सीमाएं और डेरिवेटिव

14 गणितीय तर्क

15 सांख्यिकी

16 प्रायिकता