Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.8 | Set 1

The Limits are a fundamental concept in calculus that describes the behavior of a function as its argument approaches a certain value. Understanding limits is crucial for grasping more advanced topics such as derivatives and integrals. In Class 11, Chapter 29 of RD Sharma's textbook we explore various exercises to solidify the concept of limits which is essential for students pursuing higher mathematics.

Limits

In mathematics, the limit of a function describes the value that the function approaches as the input approaches a certain point. Limits help in understanding the behavior of the functions at boundaries or points of discontinuity. They are foundational in defining derivatives and integrals thus playing a critical role in calculus. By solving problems involving the limits students can develop a deeper insight into the continuity and behavior of the functions.

Question 1. lim_x→π/2[π/2 - x].tanx

Solution:

We have,

lim_x→π/2[π/2 - x].tanx

Let us considered, y = [π/2 - x]

Here, x→π/2, y→0

= lim_y→0[y.tan(π/2 - y)]

= lim_y→0[y.{sin(π/2 - y)/cos(π/2 - y)]

= lim_y→0[y.{cosy/siny}]

= lim_y→0[y/siny].cosy

= lim_y→0[cosy]      [Since, lim_y→0[siny/y] = 1]

= 1

Question 2. lim_x→π/2[sin2x/cosx]

Solution:

We have,

lim_x→π/2[sin2x/cosx]

= lim_x→π/2[2sinx.cosx/cosx]

= 2Lim_x→π/2[sinx]

= 2

Question 3. lim_x→π/2[cos²x/(1 - sinx)]

Solution:

We have,

lim_x→π/2[cos²x/(1 - sinx)]

= lim_x→π/2[(1 - sin²x)/(1 - sinx)]

= lim_x→π/2[(1 - sinx)(1 + sinx)/(1 - sinx)]

= lim_x→π/2[(1 + sinx)]

= 1 + 1

= 2

Question 4. lim_x→π/2[(1 - sinx)/cos²x]

Solution:

We have,

lim_x→π/2[(1 - sinx)/cos²x]

= lim_x→π/2[(1 - sinx)/(1 - sin²x)]

= lim_x→π/2[(1 - sinx)/(1 - sinx)(1 + sinx)]

= lim_x→π/2[1/(1 + sinx)]

= 1/(1 + 1)

= 1/2

Question 5. lim_x→a[(cosx - cosa)/(x - a)]

Solution:

We have,

lim_x→a[(cosx - cosa)/(x - a)]

= \lim_{x\to a}[\frac{-2sin(\frac{x+a}{2})sin(\frac{x-a}{2})}{x-a}]

=-2\lim_{x\to a}[sin\frac{(x+a)}{2}]\lim_{x\to a}[\frac{sin\frac{(x-a)}{2}}{\frac{(x-a)}{2}×2}]

= -2sin[(a + a)/2] × 1 × (1/2)

= -sina

Question 6. lim_x→π/4[(1 - tanx)/(x - π/4)]

Solution:

We have,

lim_x→π/4[(1 - tanx)/(x - π/4)]

Let us considered, y = [x - π/4]

Here, x→π/4, y→0

= \lim_{y\to0}[\frac{1-tan(y+\frac{π}{4})}{y}]

= \lim_{y\to0}[\frac{1-\frac{(tany+tan\frac{π}{4})}{(1-tany.tan\frac{π}{4}}}{y}]

= \lim_{y\to0}[\frac{(1-tany-1-tany)}{y(1-tany)}]

= \lim_{y\to0}[\frac{-2tany}{y(1-tany)}]

= -2\lim_{y\to0}[\frac{tany}{y}]×\lim_{y\to0}[\frac{1}{(1-tany)}]

= -2 × 1 × [1/(1 - 0)]

= -2

Question 7. lim_x→π/2[(1 - sinx)/(π/4 - x)²]

Solution:

We have,

lim_x→π/2[(1 - sinx)/(π/4 - x)²]

Let us considered, y = [π/2 - x]

Here, x→π/2, y→0

= \lim_{y\to0}[\frac{1-sin(\frac{π}{2}-y)}{y^2}]

= lim_y→0[(1 - cosy)/y²]

= \lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{(\frac{y}{2})^2×4}]

= 2 × 1 × (1/4)

= (1/2)

Question 8. lim_x→π/3[(√3 - tanx)/(π - 3x)]

Solution:

We have,

lim_x→π/3[(√3 - tanx)/(π - 3x)]

Let us considered, y = [π/3 - x]

When, x→π/3, y→0

= \lim_{y\to0}[\frac{\sqrt3-tan(\frac{π}{3}-y)}{3y}]

= \lim_{y\to0}[\frac{\sqrt3-\frac{(tan\frac{π}{3}-tany)}{(1+tany.tan\frac{π}{3}}}{3y}]

= \lim_{y\to0}[\frac{\sqrt3-\frac{(\sqrt{3}-tany)}{(1+\sqrt{3}.tany)}}{3y}]

= \lim_{y\to0}[\frac{4tany}{3y(1+\sqrt{3}tany)}]

= \frac{4}{3}\lim_{y\to0}[\frac{tany}{y}]×\lim_{y\to0}[\frac{1}{(1+\sqrt3tany)}]

= (4/3) × 1 × [1/(1 + 0)]

= (4/3)

Question 9. lim_x→a[(asinx - xsina)/(ax² - xa²)]

Solution:

We have,

lim_x→a[(asinx - xsina)/(ax² - xa²)]

= lim_x→a[(asinx - xsina)/{ax(x - a)}]

Let us considered, y = [x - a]

When, x→a, y→0

= lim_y→0[{asin(y + a) - (y + a)sina)}/{a(y + a)y}]

= lim_y→0[(a.siny.cosa + asina.cosy - ysina - asina)/{a(y + a)y}]

= lim_y→0[{a.siny.cosa + a.sina.(cosy - 1) - y.sina}/{a(y + a)y}]

= lim_y→0[{a.siny.cosa + a.sina.2sin²(y/2) - t.sina}/{a(y + a)y}]

= lim_y→0[a.siny.cosa/a(y + a)y] - lim_y→0[2.a.sina.sin²(y/2)/a(y + a)y] + lim_y→0[y.sina/a(a + y)y]

= [(a.cosa)/a²] - [(sina)/a²] + 0

= [(a.cosa - sina)/a²]

Question 10. lim_x→π/2[{√2 - √(1 + sinx)}/cos²x]

Solution:

We have,

lim_x→π/2[{√2 - √(1 + sinx)}/cos²x]

On rationalizing the numerator, we get

= lim_x→π/2[{2 - (1 + sinx)}/cos²x{√2 + √(1 + sinx)}]

= lim_x→π/2[(1 - sinx)/(1 - sin²x){√2 + √(1 + sinx)}]

= lim_x→π/2[(1 - sinx)/(1 - sinx)(1 + sinx){√2 + √(1 + sinx)}]

= lim_x→π/2[1/(1 + sinx){√2 + √(1 + sinx)}]

= 1/{(1 + 1)(√2 + √2)}

= 1/4√2

Question 11. lim_x→π/2[{√(2 - sinx) - 1}/(π/2 - x)²]

Solution:

We have,

lim_x→π/2[{√(2 - sinx) - 1}/(π/2 - x)²]

Let us considered, y = [π/2 - x]

Here, x→π/2, y→0

=\lim_{y\to0}[\frac{\sqrt{2-sin(\frac{π}{2}-y)}-1}{y^2}]

= lim_y→0[{√(2 - cosy) - 1}/y²]

On rationalizing the numerator, we get

= lim_y→0[{(2 - cosy) - 1}/y²{√(2 - cosy) - 1}]

= lim_y→0[{1 - cosy}/y²{√(2 - cosy) - 1}]

= \lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{y^2(\sqrt{2-cosx}+1)}]

= \lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{(\frac{y}{2})^2(\sqrt{2-cosx}+1)×4}]

= 2/4(1 + 1)

= 1/4

Question 12. lim_x→π/4[(√2 - cosx - sinx)/(π/4 - x)²]

Solution:

We have,

lim_x→π/4[(√2 - cosx - sinx)/(π/4 - x)²]

= \lim_{x\to \frac{π}{2}}\frac{\sqrt2[1-(\frac{1}{\sqrt2}.cosx+\frac{1}{\sqrt2}.sinx)]}{(\frac{π}{4}-x)^2}

= \lim_{x\to \frac{π}{2}}\frac{\sqrt{2}[1-cos(\frac{π}{4}-x)]}{(\frac{π}{4}-x)^2}

= \lim_{x\to \frac{π}{2}}\frac{[2\sqrt{2}sin^2\frac{(\frac{π}{4}-x)}{2}]}{(\frac{π}{4}-x)^2}

= 2\sqrt{2}\lim_{x\to \frac{π}{2}}\frac{[sin^2\frac{(\frac{π}{4}-x)}{2}]}{\frac{(\frac{π}{4}-x)}{4}^2×4}

= 2√2/4

= (1/√2)

Question 13. lim_x→π/8[(cot4x - cos4x)/(π - 8x)³]

Solution:

We have,

lim_x→π/8[(cot4x - cos4x)/(π - 8x)³]

= lim_x→π/8[(cot4x - cos4x)/8³(π/8 - x)³]

Let us considered, (π/8 - x) = y

When x→π/8, y→0

=\lim_{x\to0}[\frac{cot(\frac{π}{2}-4x)-cos(\frac{π}{2}-4x)}{(π-8x)^3}]

= lim_x→0[(tan4x-sin4x)/8³(π/8-x)³]

= lim_x→0[(sin4x/cos4x-sin4x)/8³(π/8-x)³]

=\lim_{y\to0}[\frac{sin4y-sin4y.cos4y}{cos4y.8^3.y^3}]

=\lim_{y\to0}[\frac{sin4y(1-cos4y)}{cos4y.8^3.y^3}]

=\lim_{y\to0}[\frac{sin4y(2sin^22y)}{cos4y.8^3.y^3}]

=\frac{2}{8^3}\lim_{y\to0}\frac{sin4y}{y}×\lim_{y\to0}×\frac{sin^22y}{y^2}\lim_{y\to0}\frac{1}{cos4y}

=\frac{2}{8^3}\lim_{y\to0}\frac{sin4y}{4y}×4×\lim_{y\to0}×\frac{sin^22y}{(2y)^2}×4×\lim_{y\to0}\frac{1}{cos4y}

= (2 × 4 × 1 × 4 × 1)/(8³)

= 1/16

Question 14. lim_x→a[(cosx - cosa)/(√x - √a)]

Solution:

We have,

lim_x→a[(cosx - cosa)/(√x - √a)]

= \lim_{x\to a}[\frac{-2sin(\frac{x+a}{2})sin(\frac{x-a}{2})}{\sqrt{x}-\sqrt{a}}]

On rationalizing the denominator, we get

= \lim_{x\to a}[\frac{-2sin(\frac{x+a}{2})sin(\frac{x-a}{2})(\sqrt{x}+\sqrt{a})}{x-a}]

= -2\lim_{x\to a}[sin\frac{(x+a)}{2}]\lim_{x\to a}[\frac{sin\frac{(x-a)}{2}}{\frac{(x-a)}{2}×2}]×\lim_{x\to a}(\sqrt{x}+\sqrt{a})

= -2 × sina × 1 × (1/2) × 2√a

= -2√a.sina

Question 15. lim_x→π[{√(5 + cosx) - 2}/(π - x)²]

Solution:

We have,

lim_x→π[{√(5 + cosx) - 2}/(π - x)²]

Let us considered, y = [π - x]

When, x→π, y→0

= \lim_{y\to0}[\frac{\sqrt{5+cos(π-y)}-2}{y^2}]

= lim_y→0[{√(5 - cosy) - 2}/y²]

On rationalizing the numerator, we get

= \lim_{y\to0}[\frac{5-cosy-4}{y^2(\sqrt{5-cosy}-2)}]

= lim_y→0[{1 - cosy}/y²{√(5 - cosy)-2}]

= \lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{y^2(\sqrt{5-cosx}+2)}]

= \lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{(\frac{y}{2})^2(\sqrt{5-cosx}+2)×4}]

= 2 × (1/4) × {1/(2 + 2)}

= (1/8)

Question 16. lim_x→a[(cos√x - cos√a)/(x - a)]

Solution:

We have,

lim_x→a[(cos√x - cos√a)/(x - a)]

= \lim_{x\to a}[\frac{-2sin(\frac{\sqrt{x}+\sqrt{a}}{2})sin(\frac{\sqrt{x}-\sqrt{a}}{2})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}]

= -2\lim_{x\to a}[sin\frac{(\sqrt{x}+\sqrt{a})}{2}]\lim_{x\to a}[\frac{sin\frac{(\sqrt{x}-\sqrt{a})}{2}}{\frac{(\sqrt{x}-\sqrt{a})}{2}×2}]×\lim_{x\to a}\frac{1}{(\sqrt{x}+\sqrt{a})}

= -2sin√a × 1 × (1/2√a) × (1/2)

= -(sin√a/2√a)

Question 17. lim_x→a[(sin√x - sin√a)/(x - a)]

Solution:

We have,

lim_x→a[(sin√x - sin√a)/(x - a)]

= \lim_{x\to a}[\frac{2cos(\frac{\sqrt{x}+\sqrt{a}}{2})sin(\frac{\sqrt{x}-\sqrt{a}}{2})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}]

= 2\lim_{x\to a}[cos\frac{(\sqrt{x}+\sqrt{a})}{2}]\lim_{x\to a}[\frac{sin\frac{(\sqrt{x}-\sqrt{a})}{2}}{\frac{(\sqrt{x}-\sqrt{a})}{2}×2}]×\lim_{x\to a}\frac{1}{(\sqrt{x}+\sqrt{a})}

= 2cos√a × 1 × (1/2√a) × (1/2)

= (cos√a/2√a)

Question 18. lim_x→1[(1 - x²)/sin2πx]

Solution:

We have,

lim_x→1[(1 - x²)/sin2πx]

When, x→1, h→0

= lim_h→0[{1-(1-h)²}/sin2π(1-h)]

= lim_h→0[(2h-h²)/-sin2πh]

= lim_h→0[{h(2-h)}/sin2πh]

=-\lim_{h\to0}[\frac{(2-h)}{\frac{sin2πh}{h}}]

= -\lim_{h\to0}[\frac{(2-h)}{(\frac{sin2πh}{2πh})×2π}]

= -2/2π

= -1/π

Question 19. lim_x→π/4[{f(x) - f(π/4)}/{x - π/4}]

Solution:

We have,

lim_x→π/4[{f(x) - f(π/4)}/{x - π/4}]

When, x→π/4, h→0

= lim_h→0[{f(π/4 + h) - f(π/4)}/{π/4 + h - π/4}]

It is given that f(x) = sin2x

= lim_h→0[{sin(π/2 + 2h) - sin(π/2)}/h]

= lim_h→0[(cos2h - 1)/h]

= lim_h→0[{-2sin²h}/h]

= -2Lim_h→0[(sinh/h)²] × h

= -2 × 1 × 0

= 0

Read More:

• Limits in Calculus
• Limit Formula

Summary

Exercise 29.8 Set 1 covers evaluating limits of algebraic functions. Students learn to apply factorization, cancellation, and limit properties to find limits. Understanding limits is crucial for calculus and real-world applications. Limits help model real-world phenomena, optimize functions, and understand function behavior.