Ex 13.1, 23

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) ∴ 𝒍𝒊𝒎﷮𝐱→𝟎﷯f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮−﷯﷯ f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮−﷯﷯ 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) ∴ 𝒍𝒊𝒎﷮𝐱→𝟏﷯ f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟏﷮−﷯﷯ f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟏﷮+﷯﷯ 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→𝑎﷯ f(x) exists only if Left Hand limit = Right hand limit i.e. lim﷮ x→𝑎﷮−﷯﷯ f(x) = lim﷮ x→𝑎﷮+﷯﷯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 𝐥𝐢𝐦﷮ 𝐱→𝟎﷮−﷯﷯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 𝐥𝐢𝐦﷮ 𝐱→𝟎﷮+﷯﷯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, 𝒍𝒊𝒎﷮𝐱→𝟎﷯ f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮−﷯﷯ f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮+﷯﷯ 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) ∴ 𝒍𝒊𝒎﷮𝐱→𝟏﷯ f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟏﷮−﷯﷯ f(x) = 𝒍𝒊𝒎﷮ 𝐱→𝟏﷮+﷯﷯ f(x) = 6