Evaluate the following limits: lim h → 0 x + h - x h

lim h → 0 x + h - x h = x + 0 - x 0 = 0 0 It is indeterminate form. Applying L'Hospital's rule, differentiating w.r.t. to h , lim h → 0 x + h - x h = lim h → 0 1 2 x + h - 0 1 = lim h → 0 1 2 x + h = 1 2 x + 0 = 1 2 x

Evaluate the following limits: lim x → a x - b - a - b x 2 - a 2

lim x → a x - b - a - b x 2 - a 2 = lim x → a a - b - a - b a 2 - a 2 = 0 0 It is indeterminate form. Multiplying numerator and denominator by x - b + a - b lim x → a x - b - a - b x 2 - a 2 = lim x → a x - b - a - b x - b + a - b x + a x - a x - b + a - b = lim x → a x - b - a - b x + a x - a x - b + a - b = lim x → a x - a x + a x - a x - b + a - b = 1 a + a a - b + a - b = 1 4 a a - b

Evaluate the following limit: lim x → 0 sin 7 x 3 x

We have, lim x → 0 sin 7 x 3 x = lim x → 0 sin 7 x 7 x · 7 3 = 7 3 lim x → 0 sin 7 x 7 x = 7 3 · 1 = 7 3 ∵ lim θ → 0 sin θ θ = 1

Evaluate the following limit: lim x → 0 8 x 8 1 - cos x 2 2 - cos x 2 4 + cos x 2 2 cos x 2 4

We have, lim x → 0 8 x 8 1 - cos x 2 2 - cos x 2 4 + cos x 2 2 cos x 2 4 = lim x → 0 8 x 8 1 - cos x 2 2 - cos x 2 4 1 - cos x 2 2 = lim x → 0 8 x 8 1 - cos x 2 2 1 - cos x 2 4 = 8 lim x → 0 1 - cos x 2 2 x 2 2 × 1 - cos x 2 4 x 2 2 = 8 × 1 2 1 2 2 × 1 2 1 4 2 = 1 32 ∵ lim θ → 0 1 - cos a θ θ 2 = a 2 2

Evaluate the following limit: lim x → π 4 1 - sin 2 x π - 4 x

We have, lim x → π 4 1 - sin 2 x π - 4 x = lim x → π 4 1 - sin 2 x π - 4 x × 1 + sin 2 x 1 + sin 2 x = lim x → π 4 1 - sin 2 x 4 π 4 - x × 1 1 + sin 2 x = 1 4 lim x → π 4 1 - sin 2 x π 4 - x × lim x → π 4 1 1 + sin 2 x = 1 4 lim x → π 4 1 - sin 2 x π 4 - x × lim x → π 4 1 1 + 1 = 1 4 2 lim x → π 4 1 - sin 2 x π 4 - x Put x = π 4 - h = 1 4 2 lim h → 0 1 - sin 2 π 4 - h h = 1 4 2 lim h → 0 1 - cos 2 h h = 1 4 2 lim h → 0 1 - 1 + 2 sin 2 h h = 2 4 2 lim h → 0 sin h h = 1 4 × 1 = 1 4

Evaluate the following limit: lim x → 0 5 x - 9 x x

We have, lim x → 0 5 x - 9 x x = lim x → 0 5 x - 1 - 9 x - 1 x = lim x → 0 5 x - 1 x - lim x → 0 9 x - 1 x = log e 5 - log e 9 = log e 5 9 ∵ lim x → 0 a x - 1 x = log e a

Evaluate the following limit: lim x → ∞ 1 + a 2 x sin b 1 + a 2 x , where a ≠ 0

lim x → ∞ 1 + a 2 x sin b 1 + a 2 x = lim x → ∞ sin b 1 + a 2 - x 1 + a 2 - x = b sin b 1 + a 2 - ∞ b 1 + a 2 - ∞ = b ∵ lim x → 0 sin x x = 1

If f ' ( 3 ) = 12 , then find the value of lim h → 0 f ( 3 + h ) - f ( 3 + sinh ) h . tan 2 h

We have, lim h → 0 f ( 3 + h ) - f ( 3 + sin h ) h tan 2 h Multiply numerator and denominator by h - sin h = lim h → 0 f ( 3 + h ) - f ( 3 + sin h ) · ( h - sin h ) ( h - sin h ) h tan 2 h = lim h → 0 f ' 3 · ( h - sin h ) h tan 2 h ∵ f ' x = lim h → 0 f ( x + h ) - f ( x ) h Substitute the expansion sin h and tan h in the limit to evaluate sin h = h - h 3 3 ! + h 5 5 ! + h 7 7 ! + . . . tan h = h + h 3 3 + 2 h 5 15 + . . . lim h → 0 f ( 3 + h ) - f ( 3 + sin h ) h tan 2 h = lim h → 0 f ' 3 · h - h - h 3 3 ! + h 5 5 ! - h 7 7 ! + . . . h · h + h 3 3 + 2 h 5 15 + . . . 2 = lim h → 0 f ' 3 · h 3 3 ! - h 5 5 ! + h 7 7 ! - . . . h · h 2 1 + h 2 3 + 2 h 4 15 + . . . 2 = lim h → 0 f ' 3 · h 3 · 1 3 ! - h 2 5 ! + h 4 7 ! - . . . h · h 2 1 + h 2 3 + 2 h 4 15 + . . . 2 = lim h → 0 f ' 3 · 1 3 ! - h 2 5 ! + h 4 7 ! - . . . 1 + h 2 3 + 2 h 4 15 + . . . 2 = f ' 3 · 1 3 ! - 0 + 0 - . . . 1 + 0 + 0 + . . . 2 = f ' 3 · 1 6 = 12 · 1 6 = 2

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 25 - Limits>Exercise-1

Embibe Experts Solutions for Chapter: Limits, Exercise 1: Exercise-1

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Limits, Exercise 1: Exercise-1

Attempt the free practice questions on Chapter 25: Limits, Exercise 1: Exercise-1 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Limits, Exercise 1: Exercise-1 with Hints & Solutions

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 25 - Limits>Exercise-1>Q 4

Evaluate the following limits:
$\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 25 - Limits>Exercise-1>Q 5

Evaluate the following limits:
$\lim_{x \to a} \frac{\sqrt{x - b} - \sqrt{a - b}}{x^{2} - a^{2}}$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 25 - Limits>Exercise-1>Q 7

Evaluate the following limit:

$\lim_{x \to 0} (\frac{\sin 7 x}{3 x})$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 25 - Limits>Exercise-1>Q 8

Evaluate the following limit:
$\lim_{x \to 0} \frac{8}{x^{8}} \{1 - \cos (\frac{x^{2}}{2}) - \cos (\frac{x^{2}}{4}) + \cos (\frac{x^{2}}{2}) \cos (\frac{x^{2}}{4})\}$