# Ex 13.1, 27 - Chapter 13 Class 11 Limits and Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 13.1, 27 (Method 1) Find limx→5 f(x), where f(x) = x – 5 Finding limit at x = 5 limx→ 5+ f(x) = limx→ 5− f(x) = limx→5f(x) ∴ lim x→5+f(x) = 0 & lim x→5+f(x) = 0 Now, limx→5 f(x) = lim x→5+f(x) = lim x→5−f(x) = 0 Thus, 𝒍𝒊𝒎𝐱→𝟓 f(x) = 0 Ex 13.1, 27 (Method 2) Find limx→5 f(x), where f(x) = x – 5 We know that limx→𝑎 f(x) exists only if Left Hand limit = Right hand limit i.e. lim x→𝑎+f(x) = lim x→𝑎− f(x) We need to find limx→5 f(x) First we have to prove limx→ 5+ f(x) = limx→ 5− f(x) For 𝒍𝒊𝒎 𝐱→𝟓+f(x) f(x) = |x| – 5 So, as x tends to 5, f(x) tends to 0 ∴ lim x→5+f(x) = 0 For 𝒍𝒊𝒎 𝐱→𝟓−f(x) f(x) = |x| – 5 So, as x tends to 5, f(x) tends to 0 ∴ lim x→5−f(x) = 0 Thus, lim x→5+f(x) = 0 & lim x→5−f(x) = 0 ⇒ lim x→5+f(x) = lim x→5−f(x) = 0 Thus, limit exists & 𝒍𝒊𝒎𝐱→𝟓 f(x) = lim x→5+f(x) = lim x→5−f(x) = 0

Chapter 13 Class 11 Limits and Derivatives

Concept wise

- Limits - Defination
- Limits - 0/0 form
- Limits - x^n formula
- Limits - Of Trignometric functions
- Limits - Limit exists
- Derivatives by 1st principle - At a point
- Derivatives by 1st principle - At a general point
- Derivatives by formula - x^n formula
- Derivatives by formula - sin & cos
- Derivatives by formula - other trignometric

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.