2.6 - Review Key

# X x x2 therefore by sandwich theorem we may conclude

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Unformatted text preview: ether limx→+∞ sin(πx) and limx→+∞ 3 cos(ln(x)) actually exist and, if so, what the limit value for x2 x2 both functions is. To ﬁnd such we will use sandwich theorem. Therefore, we note that −1 ≤ sin(πx) ≤ 1 −1 x2 sin(πx) x2 ≤ We observe that lim x→∞ ≤ 1 , x2 for x = 0 −1 1 = 0 = lim 2 . 2 x→∞ x x Therefore, by sandwich theorem, we may conclude that lim x→∞ sin(πx) = 0. x2 Similarly, we note that −1 ≤ cos(ln(x)) ≤ 1 −1 x2 ≤ cos(ln(x)) x2 We observe that lim x→∞ ≤ , for x &gt; 0 1 x2 −1 1 = 0 = lim 2 . x→∞ x x2 Therefore, by sandwich theorem, we may conclude that lim x→∞ cos(ln(x)) = 0. x2 Therefore, we now note that 2 sin(πx) 2 sin(πx) lim 3 − lim + lim + 2 3−0+0 3 x→∞ x x→∞ x2 x x = x→∞ = =− . 3 cos(ln(x)) 3 cos(ln(x)) 0−5 5 −5 − lim 5 lim x→∞ x→∞ x2 x2 3− lim x→∞ 3 x→2 (x − 2)2 Recall that we can say that a limit exists if and only if the right and left-hand limits exist and are equal. Note that 3 1 1 lim = 3 lim = + = +∞ + (x − 2)2 + (x − 2)2 x→2 x→2 0 and 3 1 1 lim = 3 lim = + = +∞. − (x − 2)2 − (x − 2)2 x→2 x→2 0 3 3 3 Since lim = lim = +∞, we conclude that lim = +∞. + (x − 2)2 − (x − 2)2 x→2 (x − 2)2 x→2 x→ 2 8. lim 3 3 x→2 (x − 2)3 Recall that we can say that a limit exists if and only if the right and left-hand limits exist and are equal. Note that 3 1 1 lim = 3 lim = + = +∞ 3 3 x→2+ (x − 2) x→2+ (x − 2) 0 and 1 1 3 lim = 3 lim = − = −∞. − (x − 2)3 −...
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