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