Evaluate your nate's the Owing work is below: (24- 16) (1- cos² (F)) (i) lim 2 (sin ())(a4-8x2 + 16) (1– cos? ( )) (- 16)2 (sin ()) (ii) = lim (a4 - 822 + 16) sin () -2 (sin ()) (x4 – 8x² + 16) (a-16)2 (ii) = lim (24- 16)2 (a4-8r2 + 16) (iv) = lim sind (2-16)2 (v) = li 1. (24 - 822 + 16) (22-4) (z² + 4)² (2² – 4)² (vi) = lim 1. %3D (vii) We have a zero in the denominator since lim (x -4)2 = 0. Therefore, the limit does not exist. If your classmate made any errors in their work, state between which two lines the error(s) OFrectlvr eveluate (:) ele) ond +

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 8E
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Please explain the error and correctly evaluate
1. Evaluate your classmate's work computing the following limit. Their work is below:
(21-16)2(1-cos2 (풍))
(i) lim
232 (sin (플))(rd-8z2 + 16)
(1-cos? (풍))
(sin (풍))
(x - 16)?
(14 – 82² + 16)
(ii) = lim
sin ()
(x* – 16)²
(iii) = lim
-2 (sin ()) x4 – 8x2 + 16)
(iv) = lim
sin().
(24 - 16)2
(a4 – 8x2 + 16)
(a4- 16)2
(x4 – 8x² + 16)
(v) = lim 1
(2² – 4)²(z² + 4)²
(22 – 4)²
(vi) = lim 1.
(vii) We have a zero in the denominator since lim (x? -4)2 = 0. Therefore, the limit does not
exist.
If your classmate made any errors in their work, state between which two lines the error(s)
occurred (e.g., between line (i) and line (ii)), explain the error(s), and then correctly evaluate
(a4-16)2(1-cos2 (품))
lim
2-2 (sin ())(x4 – 8x² + 16)*
If your classmate made no errors in their work, then write 'Excellent work, that was a
challenging problem!'.
MacBook Pro
T
#3
$
%
&
3
4
5
6
7
8
Transcribed Image Text:1. Evaluate your classmate's work computing the following limit. Their work is below: (21-16)2(1-cos2 (풍)) (i) lim 232 (sin (플))(rd-8z2 + 16) (1-cos? (풍)) (sin (풍)) (x - 16)? (14 – 82² + 16) (ii) = lim sin () (x* – 16)² (iii) = lim -2 (sin ()) x4 – 8x2 + 16) (iv) = lim sin(). (24 - 16)2 (a4 – 8x2 + 16) (a4- 16)2 (x4 – 8x² + 16) (v) = lim 1 (2² – 4)²(z² + 4)² (22 – 4)² (vi) = lim 1. (vii) We have a zero in the denominator since lim (x? -4)2 = 0. Therefore, the limit does not exist. If your classmate made any errors in their work, state between which two lines the error(s) occurred (e.g., between line (i) and line (ii)), explain the error(s), and then correctly evaluate (a4-16)2(1-cos2 (품)) lim 2-2 (sin ())(x4 – 8x² + 16)* If your classmate made no errors in their work, then write 'Excellent work, that was a challenging problem!'. MacBook Pro T #3 $ % & 3 4 5 6 7 8
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