Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 13.1, 23 (Method 1) Find lim x 0 f(x) and lim x 1 f(x), where f(x) = 2x+3. 3 x+1 , x 0 x>0 Finding limit at x = 0 lim x 0 f(x) = lim x 0 + f(x) = lim x 0 f(x) f(x) = f(x) = f(x) = 3 f(x) = 2x+3. 3 x+1 , x 0 x>0 Finding limit at x = 1 lim x 1 f(x) = lim x 1 + f(x) = lim x 1 f(x) f(x) = f(x) = + f(x) = 6 Ex 13.1, 23 (Method 2) Find lim x 0 f(x) and lim x 1 f(x), where f(x) = 2x+3. 3 x+1 , x 0 x>0 Finding limit at x = 0 We know that lim x f(x) exists only if Left Hand limit = Right hand limit i.e. lim x f(x) = lim x + f(x) Similarly in question , we have to find limits First we have to prove lim x 0 f(x) = lim x 0 + f(x) For f(x) , f(x) = 2x + 3 Hence if we move value of x towards 0, value of f(x) tends towards 3 Hence, lim x 0 f(x) = 3 For + f(x) , f(x) = 3(x + 1) When we move value of x toward 0 value of f(x) tends to 3 Hence , lim x 0 + f(x) = 3 Thus lim x 0 f(x) = lim x 0 + f(x) = 3 Hence limit exists Thus, f(x) = f(x) = + f(x) = 3 f(x) = 2x+3. 3 x+1 , x 0 x>0 Finding limit at x = 1 lim x 1 f(x) = lim x 1 + f(x) = lim x 1 f(x) f(x) = f(x) = + f(x) = 6

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.