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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.
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.
Solution:
We have,
limx→π/2[π/2 - x].tanx
Let us considered, y = [π/2 - x]
Here, x→π/2, y→0
= limy→0[y.tan(π/2 - y)]
= limy→0[y.{sin(π/2 - y)/cos(π/2 - y)]
= limy→0[y.{cosy/siny}]
= limy→0[y/siny].cosy
= limy→0[cosy] [Since, limy→0[siny/y] = 1]
= 1
Solution:
We have,
limx→π/2[sin2x/cosx]
= limx→π/2[2sinx.cosx/cosx]
= 2Limx→π/2[sinx]
= 2
Solution:
We have,
limx→π/2[cos2x/(1 - sinx)]
= limx→π/2[(1 - sin2x)/(1 - sinx)]
= limx→π/2[(1 - sinx)(1 + sinx)/(1 - sinx)]
= limx→π/2[(1 + sinx)]
= 1 + 1
= 2
Solution:
We have,
limx→π/2[(1 - sinx)/cos2x]
= limx→π/2[(1 - sinx)/(1 - sin2x)]
= limx→π/2[(1 - sinx)/(1 - sinx)(1 + sinx)]
= limx→π/2[1/(1 + sinx)]
= 1/(1 + 1)
= 1/2
Solution:
We have,
limx→a[(cosx - cosa)/(x - a)]
=
=
= -2sin[(a + a)/2] × 1 × (1/2)
= -sina
Solution:
We have,
limx→π/4[(1 - tanx)/(x - π/4)]
Let us considered, y = [x - π/4]
Here, x→π/4, y→0
=
=
=
=
=
= -2 × 1 × [1/(1 - 0)]
= -2
Solution:
We have,
limx→π/2[(1 - sinx)/(π/4 - x)2]
Let us considered, y = [π/2 - x]
Here, x→π/2, y→0
=
= limy→0[(1 - cosy)/y2]
=
= 2 × 1 × (1/4)
= (1/2)
Solution:
We have,
limx→π/3[(√3 - tanx)/(π - 3x)]
Let us considered, y = [π/3 - x]
When, x→π/3, y→0
=
=
=
=
=
= (4/3) × 1 × [1/(1 + 0)]
= (4/3)
Solution:
We have,
limx→a[(asinx - xsina)/(ax2 - xa2)]
= limx→a[(asinx - xsina)/{ax(x - a)}]
Let us considered, y = [x - a]
When, x→a, y→0
= limy→0[{asin(y + a) - (y + a)sina)}/{a(y + a)y}]
= limy→0[(a.siny.cosa + asina.cosy - ysina - asina)/{a(y + a)y}]
= limy→0[{a.siny.cosa + a.sina.(cosy - 1) - y.sina}/{a(y + a)y}]
= limy→0[{a.siny.cosa + a.sina.2sin2(y/2) - t.sina}/{a(y + a)y}]
= limy→0[a.siny.cosa/a(y + a)y] - limy→0[2.a.sina.sin2(y/2)/a(y + a)y] + limy→0[y.sina/a(a + y)y]
= [(a.cosa)/a2] - [(sina)/a2] + 0
= [(a.cosa - sina)/a2]
Solution:
We have,
limx→π/2[{√2 - √(1 + sinx)}/cos2x]
On rationalizing the numerator, we get
= limx→π/2[{2 - (1 + sinx)}/cos2x{√2 + √(1 + sinx)}]
= limx→π/2[(1 - sinx)/(1 - sin2x){√2 + √(1 + sinx)}]
= limx→π/2[(1 - sinx)/(1 - sinx)(1 + sinx){√2 + √(1 + sinx)}]
= limx→π/2[1/(1 + sinx){√2 + √(1 + sinx)}]
= 1/{(1 + 1)(√2 + √2)}
= 1/4√2
Solution:
We have,
limx→π/2[{√(2 - sinx) - 1}/(π/2 - x)2]
Let us considered, y = [π/2 - x]
Here, x→π/2, y→0
=
= limy→0[{√(2 - cosy) - 1}/y2]
On rationalizing the numerator, we get
= limy→0[{(2 - cosy) - 1}/y2{√(2 - cosy) - 1}]
= limy→0[{1 - cosy}/y2{√(2 - cosy) - 1}]
=
=
= 2/4(1 + 1)
= 1/4
Solution:
We have,
limx→π/4[(√2 - cosx - sinx)/(π/4 - x)2]
=
=
=
=
= 2√2/4
= (1/√2)
Solution:
We have,
limx→π/8[(cot4x - cos4x)/(π - 8x)3]
= limx→π/8[(cot4x - cos4x)/83(π/8 - x)3]
Let us considered, (π/8 - x) = y
When x→π/8, y→0
=
= limx→0[(tan4x-sin4x)/83(π/8-x)3]
= limx→0[(sin4x/cos4x-sin4x)/83(π/8-x)3]
=
=
=
=
=
= (2 × 4 × 1 × 4 × 1)/(83)
= 1/16
Solution:
We have,
limx→a[(cosx - cosa)/(√x - √a)]
=
On rationalizing the denominator, we get
=
=
= -2 × sina × 1 × (1/2) × 2√a
= -2√a.sina
Solution:
We have,
limx→π[{√(5 + cosx) - 2}/(π - x)2]
Let us considered, y = [π - x]
When, x→π, y→0
=
= limy→0[{√(5 - cosy) - 2}/y2]
On rationalizing the numerator, we get
=
= limy→0[{1 - cosy}/y2{√(5 - cosy)-2}]
=
=
= 2 × (1/4) × {1/(2 + 2)}
= (1/8)
Solution:
We have,
limx→a[(cos√x - cos√a)/(x - a)]
=
=
= -2sin√a × 1 × (1/2√a) × (1/2)
= -(sin√a/2√a)
Solution:
We have,
limx→a[(sin√x - sin√a)/(x - a)]
=
=
= 2cos√a × 1 × (1/2√a) × (1/2)
= (cos√a/2√a)
Solution:
We have,
limx→1[(1 - x2)/sin2πx]
When, x→1, h→0
= limh→0[{1-(1-h)2}/sin2π(1-h)]
= limh→0[(2h-h2)/-sin2πh]
= limh→0[{h(2-h)}/sin2πh]
=
=
= -2/2π
= -1/π
Solution:
We have,
limx→π/4[{f(x) - f(π/4)}/{x - π/4}]
When, x→π/4, h→0
= limh→0[{f(π/4 + h) - f(π/4)}/{π/4 + h - π/4}]
It is given that f(x) = sin2x
= limh→0[{sin(π/2 + 2h) - sin(π/2)}/h]
= limh→0[(cos2h - 1)/h]
= limh→0[{-2sin2h}/h]
= -2Limh→0[(sinh/h)2] × h
= -2 × 1 × 0
= 0
Read More:
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.
