MATH 12002 - CALCULUS I § 1.4: Calculating Limits - More Examples Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7
Limits & Radicals Example 1 √ x + 1 − 2 Evaluate lim . x − 3 x → 3 Observe first that when x = 3, both the numerator and denominator are 0. This indicates that x − 3 is a factor of the numerator. To use “cancellation” to evaluate the limit, we need the other factor. When evaluating limits involving radicals, it is often helpful to “rationalize” the numerator or denominator. In this case, we multiply the numerator and denominator by the conjugate , √ x + 1 + 2, of the numerator and use the fact that ( a − b )( a + b ) = a 2 − b 2 to obtain ( √ x + 1 − 2)( √ x + 1 + 2) = ( √ x + 1) 2 − 2 2 = ( x + 1) − 4 = x − 3. [Continued → ] D.L. White (Kent State University) 2 / 7
Limits & Radicals We therefore have √ x + 1 − 2 ( √ x + 1 − 2)( √ x + 1 + 2) lim = lim ( x − 3)( √ x + 1 + 2) x − 3 x → 3 x → 3 ( x + 1) − 4 = lim ( x − 3)( √ x + 1 + 2) x → 3 x − 3 = lim ( x − 3)( √ x + 1 + 2) x → 3 1 = lim √ x + 1 + 2 , since x � = 3 x → 3 1 = √ 3 + 1 + 2 , by Limit Laws 1 2 + 2 = 1 1 = = √ 4 . 4 + 2 D.L. White (Kent State University) 3 / 7
Limits & Sums We know that if lim x → a f ( x ) and lim x → a g ( x ) both exist, then x → a [ f ( x ) + g ( x )] = lim lim x → a f ( x ) + lim x → a g ( x ). But it is also possible for lim x → a [ f ( x ) + g ( x )] to exist even if lim x → a f ( x ) and lim x → a g ( x ) do not. Example 2 x 2 − 2 x + 7 � 3 � Evaluate lim x − 1 + . ( x − 1)( x − 3) x → 1 x 2 − 2 x + 7 3 Note that neither lim x − 1 nor lim ( x − 1)( x − 3) exists, x → 1 x → 1 because both denominators are 0 when x = 1, while neither numerator is 0. To evaluate the limit, we will need to combine the fractional expressions. [Continued → ] D.L. White (Kent State University) 4 / 7
Limits & Sums We have x 2 − 2 x + 7 x 2 − 2 x + 7 � 3 � � 3( x − 3) � lim x − 1 + = lim ( x − 1)( x − 3) + ( x − 1)( x − 3) ( x − 1)( x − 3) x → 1 x → 1 3( x − 3) + x 2 − 2 x + 7 = lim ( x − 1)( x − 3) x → 1 3 x − 9 + x 2 − 2 x + 7 = lim ( x − 1)( x − 3) x → 1 x 2 + x − 2 = lim ( x − 1)( x − 3) x → 1 ( x − 1)( x + 2) = lim ( x − 1)( x − 3) x → 1 x + 2 = lim x − 3 , since x � = 1 x → 1 = 1 + 2 1 − 3 = 3 − 2 = − 3 2 . D.L. White (Kent State University) 5 / 7
Squeeze Theorem Some limits may be difficult to compute directly, but may be closely related to other limits that are easy to compute. In these cases, the Squeeze Theorem may be useful. Squeeze Theorem If f ( x ) � g ( x ) � h ( x ) for all x in an open interval around x = a (except possibly at x = a), and lim x → a f ( x ) = lim x → a h ( x ) = L, then lim x → a g ( x ) = L. Example 3 x → 0 (sin 2 x ) cos 1 Evaluate lim x . Observe that as x → 0, sin 2 x → sin 2 0 = 0, but lim x → 0 cos 1 x does not exist. Therefore, we cannot use limit laws to evaluate the limit of the product. We can, however, use the Squeeze Theorem. [Continued → ] D.L. White (Kent State University) 6 / 7
Squeeze Theorem We know that − 1 � cos 1 x � 1. Since sin 2 x � 0 for all x , we can multiply each part of this inequality by sin 2 x to obtain − sin 2 x � (sin 2 x ) cos 1 x � sin 2 x . x → 0 sin 2 x = 0, and so lim x → 0 ( − sin 2 x ) = 0. As we observed above, lim Therefore, (sin 2 x ) cos 1 x is squeezed between − sin 2 x and sin 2 x , both of which approach 0 as x approaches 0. x → 0 (sin 2 x ) cos 1 Hence the Squeeze Theorem implies lim x = 0. D.L. White (Kent State University) 7 / 7
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