11 Maths NcertSolutions Chapter 13 2 Supplementary

Mathematics (www.tiwariacademy.com)

(Chapter – 13) (Limits and Derivatives) (Class – XI)

Exercise 13.2 (Supplementary) 𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑙𝑖𝑚𝑖𝑡𝑠, 𝑖𝑓 𝑒𝑥𝑖𝑠𝑡. Question 1: Answer 1:

𝑒 4𝑥 −1 𝑥→0 𝑥

lim

𝑒 4𝑥 −1 𝑥→0 𝑥

lim

𝑒 4𝑥 −1 × 𝑥→0 4𝑥

4

𝑒 𝑦 −1 × 𝑦→0 𝑦

4

= lim

= lim = 1×4

[ 𝑊ℎ𝑒𝑟𝑒 𝑦 = 4𝑥 ] 𝑒 𝑦 −1 𝑦→0 𝑦

= 1]

𝑒 𝑥 −1 𝑥→0 𝑥

= 1]

[Using lim

=4

Question 2:

𝑒 2+𝑥 −𝑒 2 𝑥 𝑥→0

Answer 2:

𝑒 2+𝑥 −𝑒 2 𝑥 𝑥→0

lim lim

𝑒 2 (𝑒 𝑥 −1) 𝑥 𝑥→0

= lim

= 𝑒2 × 1

[Using lim

= 𝑒2

Question 3:

𝑒𝑥 − 𝑒5 𝑥→5 𝑥 − 5

Answer 3:

𝑒𝑥 − 𝑒5 𝑥→5 𝑥−5

lim

lim

𝑃𝑢𝑡 𝑥 = 5 + ℎ, 𝑡ℎ𝑒𝑛 𝑎𝑠 𝑥 → 5 ⟹ ℎ ⟶ 0. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒

1 www.tiwariacademy.com

Mathematics (www.tiwariacademy.com)

(Chapter – 13) (Limits and Derivatives) (Class – XI)

𝑒𝑥 − 𝑒5 𝑥→5 𝑥−5

𝑒 5+ℎ −𝑒 5 ℎ ℎ→0

lim

= lim

= lim

𝑒 5 (𝑒 ℎ −1)

ℎ→0

ℎ 𝑒 ℎ −1 ℎ→0 ℎ

= 𝑒5 × 1

[Using lim

= 1]

= 𝑒5

Question 4:

𝑒 𝑠𝑖𝑛 𝑥 −1 𝑥 𝑥→0

Answer 4:

𝑒 𝑠𝑖𝑛 𝑥 −1 𝑥 𝑥→0

lim lim

𝑒 𝑠𝑖𝑛 𝑥 −1 𝑥 𝑥→0

× sin 𝑥

𝑒 𝑠𝑖𝑛 𝑥 −1 𝑥→0 sin 𝑥

×

= lim = lim

𝑒 𝑦 −1 𝑦→0 𝑦

= lim

=1 × 1

sin 𝑥

sin 𝑥 𝑥

× lim

𝑥→0

sin 𝑥 𝑥

[𝑊ℎ𝑒𝑟𝑒 𝑦 = sin 𝑥] 𝑒 𝑦 −1 𝑦→0 𝑦

[Using lim

=1

Question 5:

𝑒𝑥 − 𝑒3 𝑥→3 𝑥 − 3

Answer 5:

𝑒𝑥 − 𝑒3 𝑥→3 𝑥−3

lim

lim

𝑃𝑢𝑡 𝑥 = 3 + ℎ, 𝑡ℎ𝑒𝑛 𝑎𝑠 𝑥 → 3 ⟹ ℎ ⟶ 0. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒

2 www.tiwariacademy.com

sin 𝑥 𝑥→0 𝑥

= 1 𝑎𝑛𝑑 lim

= 1]

Mathematics (www.tiwariacademy.com)

(Chapter – 13) (Limits and Derivatives) (Class – XI) 𝑒𝑥 − 3 𝑥→3 𝑥−3

lim

= lim

𝑒 3+ℎ −𝑒 3 ℎ ℎ→0

= lim

𝑒 3 (𝑒 ℎ −1) ℎ

ℎ→0

= 𝑒3 × 1

𝑒 ℎ −1 ℎ→0 ℎ

= 1]

𝑒 𝑥 −1 𝑥→0 𝑥

= 1 𝑎𝑛𝑑 lim

[Using lim

= 𝑒3

Question 6:

𝑥 (𝑒 𝑥 −1) 𝑥→0 1−cos 𝑥

Answer 6:

𝑥 (𝑒 𝑥 −1) 𝑥→0 1−cos 𝑥

lim lim

𝑥 (𝑒 𝑥 −1) 1+cos 𝑥 × 1+cos 𝑥 𝑥→0 1−cos 𝑥

= lim = lim

𝑥→0

= lim

𝑥→0

= lim

𝑥→0

(𝑒 𝑥 −1) 𝑥 (𝑒 𝑥 −1) 𝑥 (𝑒 𝑥 −1) 𝑥

×

1+cos 𝑥 1

𝑥

×𝑥

×

𝑥2 1− 𝑐𝑜𝑠2 𝑥 𝑥2 𝑥→0 𝑠𝑖𝑛2 𝑥

1+cos 𝑥 1 𝑥→0

× lim

1+cos 𝑥 1 𝑥→0

× lim

× lim × lim 1

= 1 × (1 + 1) × 11

1

2 𝑥→0 ( sin 𝑥) 𝑥

[Using lim

=2

Question 7: Answer 7:

𝑙𝑜𝑔𝑒 (1+2𝑥) 𝑥 𝑥→0

lim lim

𝑥→0

𝑙𝑜𝑔𝑒 (1+2𝑥) 𝑥

𝑙𝑜𝑔𝑒 (1+2𝑥) ×2 2𝑥 𝑥→0

= lim

3 www.tiwariacademy.com

𝑥→0

sin 𝑥 𝑥

= 1]

Mathematics (www.tiwariacademy.com)

(Chapter – 13) (Limits and Derivatives) (Class – XI) 𝑙𝑜𝑔𝑒 (1+𝑦) × 𝑦 𝑦→0

= lim

[𝑊ℎ𝑒𝑟𝑒 𝑦 = 2𝑥]

2

=1×2

[𝑈𝑠𝑖𝑛𝑔 lim

𝑦→0

𝑙𝑜𝑔𝑒 (1+𝑦) 𝑦

= 1]

=2

Question 8: Answer 8:

lim

𝑙𝑜𝑔 (1+𝑥 3 ) 𝑠𝑖𝑛3 𝑥

𝑥→0

lim

𝑙𝑜𝑔 (1+𝑥 3 ) 𝑠𝑖𝑛3 𝑥

𝑥→0

𝑙𝑜𝑔 (1+𝑥 3 )

= lim

𝑠𝑖𝑛3 𝑥

𝑥→0

= lim

𝑙𝑜𝑔 (1+𝑥 3 ) 𝑥3

𝑥→0

= lim

𝑦→0

𝑥3

× 𝑥3 ×

𝑥3 𝑠𝑖𝑛3 𝑥

𝑙𝑜𝑔 (1+𝑦) 1 × lim sin 𝑥 3 𝑦 𝑥→0 ( )

[𝑊ℎ𝑒𝑟𝑒 𝑦 = 𝑥 3 ]

𝑥

1

= 1 × 13

𝑙𝑜𝑔 (1+𝑦) 𝑦 𝑦→0

[𝑈𝑠𝑖𝑛𝑔 lim

=1

4 www.tiwariacademy.com

= 1 𝑎𝑛𝑑 lim

𝑥→0

sin 𝑥 𝑥

= 1]