Problem Set 02s

# Lim solution given 0 we need 0 such that 0

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Unformatted text preview: the values of ⎡ྎ 2 ⎛ྎ 2π x ⎢ྎ1 + sin ⎜ྎ ⎟ྏ⎥ྏ = lim x + lim+ x ྎsin ⎜ྎ x →0 ⎝ྎ x ⎠ྏ⎦ྏ x→0+ ⎝ྎ x ⎣ྏ ⎣ྏ ⎞ྏ⎤ྏ ⎟ྏ⎥ྏ ⎠ྏ⎦ྏ And we know that lim x = 0 x →0+ We have to use squeeze theorem to solve for ⎡ྎ ⎛ྎ 2π lim+ x ⎢ྎsin 2 ⎜ྎ x →0 ⎝ྎ x ⎣ྏ We know that And ⎛ྎ 2π 0 ≤ sin 2 ⎜ྎ ⎝ྎ x ⎛ྎ 2π 0 ≤ sin 2 ⎜ྎ ⎝ྎ x ⎞ྏ⎤ྏ ⎟ྏ⎥ྏ ⎠ྏ⎦ྏ ⎞ྏ ⎟ྏ ≤ 1 ⎠ྏ ⎞ྏ ⎟ྏ ≤ x ⎠ྏ Using the squeeze theorem lim h( x) = lim+ x = 0 x →0+ x →0 And given that k ( x) ≤ g ( x) ≤ h( x) lim+ g ( x) = 0 x →0 2 Hung Nguyen Problem Set 02 Solution GE207C 2.3 Limits Laws/Squeeze Theorem 2.4 Precise Definition of Limit x−2 < 10 4 −2 = 3 3 For both of these conditions to be satisfied at once, we choose the more restrictive condition, so δ= 4 . 7 5. Prove the statement by using the ε and δ definition of limit x3 = x→3 5 5 a. lim Solution: Given ε > 0 , we need δ > 0 such that 0 < x − 3 < δ , then f ( x) − 3 < ε 5 3 x3 1 < ε...
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