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Ex 13.1, 32 - If f(x) = {mx^2 + n, nx + m, nx^3 + m. For what m and n

Ex 13.1, 32 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.1, 32 - Chapter 13 Class 11 Limits and Derivatives - Part 3 Ex 13.1, 32 - Chapter 13 Class 11 Limits and Derivatives - Part 4

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Ex 13.1, 32 If f(x) = {█(mx2+n, x<[email protected]+m 0≤x≤[email protected]+m, x>1)┤ . For what integers m and n does lim┬(x→0) f(x) and lim┬(x→1) f(x) exist? Given limit exists at x = 0 and x = 1 At x = 0 Limit exists at x = 0 if Left hand limit = Right hand limit If f(x) = {█(mx2+n, x<[email protected]+m 0≤x≤[email protected]+m, x>1)┤ LHL at x → 0 lim┬(x→0^− ) f(x) = lim┬(h→0) f(0 − h) = lim┬(h→0) f(−h) = lim┬(h→0) mh^2+n = m(0)2 + n = n RHL at x → 0 lim┬(x→0^+ ) f(x) = lim┬(h→0) f(0 + h) = lim┬(h→0) f (h) = lim┬(h→0) 𝑛ℎ+𝑚 = n(0) + m = m Since LHL = RHL ∴ m = n So, lim┬(x→0) f(x) exists if m = n Now, Limit exists at x = 1 Thefore, Left hand limit = Right hand limit f(x) = {█(mx2+n, x<[email protected]+m 0≤x≤[email protected]+m, x>1)┤ LHL at x → 1 lim┬(x→1^− ) f(x) = lim┬(h→0) f(1 − h) = lim┬(h→0) 𝑛(1−ℎ)+𝑚 = n(1 – 0) + m = n + m RHL at x → 1 lim┬(x→1^+ ) f(x) = lim┬(h→0) f(1 + h) = lim┬(h→0) 𝑛(1+ℎ)^3+𝑚 = n(1 + 0)3 + m = n + m Since LHL = RHL m + n = m + n But, this is always true So, lim┬(x→1) f(x) exists at all integral values of m & n

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.