UProblem 5: (a) Evaluate:6 x cos(5a+ 2)limx-09x- 4(Hint: The squeeze theorem may be helpful.)(b) Prove that:1-cos(a)21limBonus: Billy Bob Joe says that the following limit:-2x7limx5 2-25x2-3x10does not exist, since the numerator is non-zero and the denominator is zero in both terms. Explainwhy Billy is wrong, and calculate the value of the limit.

Question
Asked Nov 1, 2019
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U
Problem 5: (a) Evaluate:
6 x cos(5a+ 2)
lim
x-0
9x- 4
(Hint: The squeeze theorem may be helpful.)
(b) Prove that:
1-cos(a)
2
1
lim
Bonus: Billy Bob Joe says that the following limit:
-2x
7
lim
x5 2-25
x2-3x10
does not exist, since the numerator is non-zero and the denominator is zero in both terms. Explain
why Billy is wrong, and calculate the value of the limit.
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U Problem 5: (a) Evaluate: 6 x cos(5a+ 2) lim x-0 9x- 4 (Hint: The squeeze theorem may be helpful.) (b) Prove that: 1-cos(a) 2 1 lim Bonus: Billy Bob Joe says that the following limit: -2x 7 lim x5 2-25 x2-3x10 does not exist, since the numerator is non-zero and the denominator is zero in both terms. Explain why Billy is wrong, and calculate the value of the limit.

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Expert Answer

Step 1
  • Given limit
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6+xcos(5x2 lim х—0 9х-4 put the value of x-0 and check the limit exist or not б+0xсos(5x01+2) 6+0xcosoo 5 +2cos+2=cosso cos 9x0-4 -4 value ofcosoo lie between -1 to +1 |so, when multiply by 0 it give zero value 6+0 6 -4 -4 2

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Step 2

Hence limiting value is given as

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6+xcos(5x+2) lim x-0 9x-4 2

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Step 3
  • Given limit
...
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1-cosx lim x0

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Calculus