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Unformatted text preview: Math 341 Homework # 5 P79. 1, 3, 5, 8. 2− 1. Deﬁne f : (−2, 0) → R by f (x) = x +24 . Prove that f has a limit at −2, x and ﬁnd it. Proof: Note that (x + 2)(x − 2) = x − 2. f (x) = x+2 Guess the limit will be −4. ∀ > 0, ∃ δ = > 0, ∀ 0 < x + 2 < δ , we have f (x) + 4 = x − 2 + 4 = x + 2 < δ = . Thus
x→−2 lim f (x) = −4. 3. Give an example of a function f : (0, 1) → R that has a limit at every 1 point of (0, 1) except 2 . Use the deﬁnition of limit of a function to justify the example. Proof: Let 1 x> 1 2 f (x) = −1 x ≤ 1 . 2 For x0 = 1 , ∀ > 0, ∃ δ = 2 or x, x0 < 1 . Thus 2
1 For x0 = 2 , suppose x→x0 1 2 − x0 > 0, ∀ 0 < x − x0  < δ , we have x, x0 > 1 2 f (x) − f (x0 ) = 0 < . lim f (x) = L.
1 2 Then, for = 1, ∃δ > 0, ∀ 0 < x − < δ , we have f (x) − L < 1. Take x > 1 , then 1 − L < 1 and hence, 0 < L < 2. 2 1 Take x < 2 , then  − 1 − L < 1 and hence, −2 < L < 0. Thus, L < 0 < L which is not possible. This proves that f does not have a limit at 1 . 2 1 5. Suppose f : D → R with x0 an accumulation point of D. Assume L1 and L2 are limits of f at x0 . Prove L1 = L2 . Proof: Suppose L1 = L2 . We may assume L1 > L2 . Since limx→x0 f (x) = L1 , ∀ > 0, ∃ δ1 > 0, ∀ 0 < x − x0  < δ1 , we have f (x) − L1  < and hence, L1 − < f (x) < L1 + . Similarly, since limx→x0 f (x) = L1 , ∀ > 0, ∃ δ2 > 0, ∀ 0 < x − x0  < δ2 , we have L2 − < f (x) < L2 + . Take δ = min(δ1 , δ2 ) and =
L1 −L2 . 2 Then, for 0 < x − x0  < δ , f (x) < L2 + = L1 − < f (x). This is impossible and hence, L1 = L2 . 8. Deﬁne f : (0, 1) → R by f (x) = Proof: Note that f (x) =
x3 −x2 +x−1 . x−1 Prove that f has a limit at 1. x2 (x − 1) + x − 1 = x2 + 1 . x−1 We guess the limit is 2. ∀ > 0, ∃ δ = min 1, 3 > 0, ∀ 0 < x − 1 < δ , we have x ∈ (1 − δ, 1 + δ ) ⊂ (0, 2), so 1 < x + 1 < 3, and hence f (x) − 2 = x2 + 1 − 2 = x2 − 1 = x + 1x − 1 < 3δ ≤ . Therefore
x→1 lim f (x) = 2. 2 ...
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 Spring '07
 thieme
 Math, Calculus, Limit

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