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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 Winter '06
 McAdam
 Calculus, Differential Calculus, Limits, Limit, lim, lim T

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